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Topic 5 —Calculus

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Topic 5 —Calculus

16N.1.AHL.TZ0.H_9a

Question

A curve has equation $3 x - 2 y^{2} e^{x - 1} = 2$ .

Find an expression for $\frac{d y}{d x}$ in terms of $x$ and $y$ .

[5]

Find the equations of the tangents to this curve at the points where the curve intersects the line $x = 1$ .

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to differentiate implicitly M1

$3 - (4 y \frac{d y}{d x} + 2 y^{2}) e^{x - 1} = 0$ A1A1A1

Note: Award A1 for correctly differentiating each term.

$\frac{d y}{d x} = \frac{3 ∙ e^{1 - x} - 2 y^{2}}{4 y}$ A1

Note: This final answer may be expressed in a number of different ways.

[5 marks]

$3 - 2 y^{2} = 2 \Rightarrow y^{2} = \frac{1}{2} \Rightarrow y = \pm \sqrt{\frac{1}{2}}$ A1

$\frac{d y}{d x} = \frac{3 - 2 ∙ \frac{1}{2}}{\pm 4 \sqrt{\frac{1}{2}}} = \pm \frac{\sqrt{2}}{2}$ M1

at $(1, \sqrt{\frac{1}{2}})$ the tangent is $y - \sqrt{\frac{1}{2}} = \frac{\sqrt{2}}{2} (x - 1)$ and A1

at $(1, - \sqrt{\frac{1}{2}})$ the tangent is $y + \sqrt{\frac{1}{2}} = - \frac{\sqrt{2}}{2} (x - 1)$ A1

Note: These equations simplify to $y = \pm \frac{\sqrt{2}}{2} x$ .

Note: Award A0M1A1A0 if just the positive value of $y$ is considered and just one tangent is found.

[4 marks]

Examiners report

[N/A]

16N.1.AHL.TZ0.H_11c

Question

Consider the function $f$ defined by $f (x) = e^{x} \sin x, 0 ⩽ x ⩽ π$ .

The curvature at any point $(x, y)$ on a graph is defined as $κ = \frac{| \frac{d^{2} y}{d x^{2}} |}{{(1 + {(\frac{d y}{d x})}^{2})}^{\frac{3}{2}}}$ .

Show that the function $f$ has a local maximum value when $x = \frac{3 π}{4}$ .

[2]

Find the $x$ -coordinate of the point of inflexion of the graph of $f$ .

[2]

Sketch the graph of $f$ , clearly indicating the position of the local maximum point, the point of inflexion and the axes intercepts.

[3]

Find the area of the region enclosed by the graph of $f$ and the $x$ -axis.

The curvature at any point $(x, y)$ on a graph is defined as $κ = \frac{| \frac{d^{2} y}{d x^{2}} |}{{(1 + {(\frac{d y}{d x})}^{2})}^{\frac{3}{2}}}$ .

[6]

Find the value of the curvature of the graph of $f$ at the local maximum point.

[3]

Find the value $κ$ for $x = \frac{π}{2}$ and comment on its meaning with respect to the shape of the graph.

[2]

Markscheme

$\frac{d y}{d x} = e^{\frac{3 π}{4}} (\sin \frac{3 π}{4} + \cos \frac{3 π}{4}) = 0$ R1

$\frac{d^{2} y}{d x^{2}} = 2 e^{\frac{3 π}{4}} \cos \frac{3 π}{4} < 0$ R1

hence maximum at $x = \frac{3 π}{4}$ AG

[2 marks]

$\frac{d^{2} y}{d x^{2}} = 0 \Rightarrow 2 e^{x} \cos x = 0$ M1

$\Rightarrow x = \frac{π}{2}$ A1

Note: Award M1A0 if extra zeros are seen.

[2 marks]

N16/5/MATHL/HP1/ENG/TZ0/11.e/M

correct shape and correct domain A1

max at $x = \frac{3 π}{4}$ , point of inflexion at $x = \frac{π}{2}$ A1

zeros at $x = 0$ and $x = π$ A1

Note: Penalize incorrect domain with first A mark; allow FT from (d) on extra points of inflexion.

[3 marks]

EITHER

$\int_{0}^{x} e^{x} \sin x d x = [e^{x} \sin x]_{0}^{π} - \int_{0}^{π} e^{x} \cos x d x$ M1A1

$\int_{0}^{π} e^{x} \sin x d x = [e^{x} \sin x]_{0}^{π} - ([e^{x} \cos x]_{0}^{x} + \int_{0}^{π} e^{x} \sin x d x)$ A1

$\int_{0}^{π} e^{x} \sin x d x = [- e^{x} \cos x]_{0}^{π} + \int_{0}^{π} e^{x} \cos x d x$ M1A1

$\int_{0}^{π} {e^{x} \sin x d x = [- e^{x} \cos x]}_{0}^{π} + ([e^{x} \sin x]_{0}^{π} - \int_{0}^{π} e^{x} \sin x d x)$ A1

THEN

$\int_{0}^{π} e^{x} \sin x d x = \frac{1}{2} ([e^{x} \sin x]_{0}^{x} - [e^{x} \cos x]_{0}^{x})$ M1A1

$\int_{0}^{π} e^{x} \sin x d x = \frac{1}{2} (e^{x} + 1)$ A1

[6 marks]

$\frac{d y}{d x} = 0$ (A1)

$\frac{d^{2} y}{d x^{2}} = 2 e^{\frac{3 π}{4}} \cos \frac{3 π}{4} = - \sqrt{2} e^{\frac{3 π}{4}}$ (A1)

$κ = \frac{| - \sqrt{2} e^{\frac{3 π}{4}} |}{1} = \sqrt{2} e^{\frac{3 π}{4}}$ A1

[3 marks]

$κ = 0$ A1

the graph is approximated by a straight line R1

[2 marks]

Examiners report

[N/A]

17M.1.AHL.TZ2.H_4a

Question

A particle moves along a straight line. Its displacement, $s$ metres, at time $t$ seconds is given by $s = t + \cos 2 t, t ⩾ 0$ . The first two times when the particle is at rest are denoted by $t_{1}$ and $t_{2}$ , where $t_{1} < t_{2}$ .

Find $t_{1}$ and $t_{2}$ .

[5]

Find the displacement of the particle when $t = t_{1}$

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$s = t + \cos 2 t$

$\frac{d s}{d t} = 1 - 2 \sin 2 t$ M1A1

$= 0$ M1

$\Rightarrow \sin 2 t = \frac{1}{2}$

$t_{1} = \frac{π}{12} (s), t_{2} = \frac{5 π}{12} (s)$ A1A1

Note: Award A0A0 if answers are given in degrees.

[5 marks]

$s = \frac{π}{12} + \cos \frac{π}{6} (s = \frac{π}{12} + \frac{\sqrt{3}}{2} (m))$ A1A1

[2 marks]

Examiners report

[N/A]

17M.1.AHL.TZ2.H_6a

Question

Using the substitution $x = \tan θ$ show that $\int_{0}^{1} \frac{1}{{(x^{2} + 1)}^{2}} d x = \int_{0}^{\frac{π}{4}} \cos^{2} θ d θ$ .

[4]

Hence find the value of $\int_{0}^{1} \frac{1}{{(x^{2} + 1)}^{2}} d x$ .

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

let $x = \tan θ$

$\Rightarrow \frac{d x}{d θ} = \sec^{2} θ$ (A1)

$\int \frac{1}{{(x^{2} + 1)}^{2}} d x = \int \frac{\sec^{2} θ}{{(\tan^{2} θ + 1)}^{2}} d θ$ M1

Note: The method mark is for an attempt to substitute for both $x$ and $d x$ .

$= \int \frac{1}{\sec^{2} θ} d θ$ (or equivalent) A1

when $x = 0, θ = 0$ and when $x = 1, θ = \frac{π}{4}$ M1

$\int_{0}^{\frac{π}{4}} \cos^{2} θ d θ$ AG

[4 marks]

$(\int_{0}^{1} \frac{1}{{(x^{2} + 1)}^{2}} d x = \int_{0}^{\frac{π}{4}} \cos^{2} θ d θ) = \frac{1}{2} \int_{0}^{\frac{π}{4}} (1 + \cos 2 θ) d θ$ M1

$= \frac{1}{2} {[θ + \frac{\sin 2 θ}{2}]}_{0}^{\frac{π}{4}}$ A1

$= \frac{π}{8} + \frac{1}{4}$ A1

[3 marks]

Examiners report

[N/A]

17M.1.AHL.TZ1.H_9

Question

Find $\int \arcsin x d x$

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt at integration by parts with $u = \arcsin x$ and $v^{'} = 1$ M1

$\int \arcsin x d x = x \arcsin x - \int \frac{x}{\sqrt{1 - x^{2}}} d x$ A1A1

Note: Award A1 for $x \arcsin x$ and A1 for $- \int \frac{x}{\sqrt{1 - x^{2}}} d x$ .

solving $\int \frac{x}{\sqrt{1 - x^{2}}} d x$ by substitution with $u = 1 - x^{2}$ or inspection (M1)

$\int \arcsin x d x = x \arcsin x + \sqrt{1 - x^{2}} + c$ A1

[5 marks]

Examiners report

[N/A]

17M.1.AHL.TZ2.H_9b

Question

Consider the function $f$ defined by $f (x) = x^{2} - a^{2}, x \in R$ where $a$ is a positive constant.

The function $g$ is defined by $g (x) = x \sqrt{f (x)}$ for $| x | > a$ .

Showing any $x$ and $y$ intercepts, any maximum or minimum points and any asymptotes, sketch the following curves on separate axes.

$y = f (x)$ ;

[2]

a.i.

Showing any $x$ and $y$ intercepts, any maximum or minimum points and any asymptotes, sketch the following curves on separate axes.

$y = \frac{1}{f (x)}$ ;

[4]

a.ii.

Showing any $x$ and $y$ intercepts, any maximum or minimum points and any asymptotes, sketch the following curves on separate axes.

$y = | \frac{1}{f (x)} |$ .

[2]

a.iii.

Find $\int f (x) \cos x d x$ .

[5]

By finding $g^{'} (x)$ explain why $g$ is an increasing function.

[4]

Markscheme

M17/5/MATHL/HP1/ENG/TZ2/09.a.i/M

A1 for correct shape

A1 for correct $x$ and $y$ intercepts and minimum point

[2 marks]

a.i.

M17/5/MATHL/HP1/ENG/TZ2/09.a.ii/M

A1 for correct shape

A1 for correct vertical asymptotes

A1 for correct implied horizontal asymptote

A1 for correct maximum point

[??? marks]

a.ii.

M17/5/MATHL/HP1/ENG/TZ2/09.a.iii/M

A1 for reflecting negative branch from (ii) in the $x$ -axis

A1 for correctly labelled minimum point

[2 marks]

a.iii.

EITHER

attempt at integration by parts (M1)

$\int (x^{2} - a^{2}) \cos x d x = (x^{2} - a^{2}) \sin x - \int 2 x \sin x d x$ A1A1

$= (x^{2} - a^{2}) \sin x - 2 [- x \cos x + \int \cos x d x]$ A1

$= (x^{2} - a^{2}) \sin x + 2 x \cos - 2 \sin x + c$ A1

$\int (x^{2} - a^{2}) \cos x d x = \int x^{2} \cos x d x - \int a^{2} \cos x d x$

attempt at integration by parts (M1)

$\int x^{2} \cos x d x = x^{2} \sin x - \int 2 x \sin x d x$ A1A1

$= x^{2} \sin x - 2 [- x \cos x + \int \cos x d x]$ A1

$= x^{2} \sin x + 2 x \cos x - 2 \sin x$

$- \int a^{2} \cos x d x = - a^{2} \sin x$

$\int (x^{2} - a^{2}) \cos x d x = (x^{2} - a^{2}) \sin x + 2 x \cos x - 2 \sin x + c$ A1

[5 marks]

$g (x) = x (x^{2} - a^{2})^{\frac{1}{2}}$

$g^{'} (x) = (x^{2} - a^{2})^{\frac{1}{2}} + \frac{1}{2} x (x^{2} - a^{2})^{- \frac{1}{2}} (2 x)$ M1A1A1

Note: Method mark is for differentiating the product. Award A1 for each correct term.

$g^{'} (x) = (x^{2} - a^{2})^{\frac{1}{2}} + x^{2} (x^{2} - a^{2})^{- \frac{1}{2}}$

both parts of the expression are positive hence $g^{'} (x)$ is positive R1

and therefore $g$ is an increasing function (for $| x | > a$ ) AG

[4 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

a.iii.

[N/A]

17M.1.AHL.TZ1.H_11b

Question

Consider the function $f (x) = \frac{1}{x^{2} + 3 x + 2}, x \in R, x \neq - 2, x \neq - 1$ .

Express $x^{2} + 3 x + 2$ in the form $(x + h)^{2} + k$ .

[1]

a.i.

Factorize $x^{2} + 3 x + 2$ .

[1]

a.ii.

Sketch the graph of $f (x)$ , indicating on it the equations of the asymptotes, the coordinates of the $y$ -intercept and the local maximum.

[5]

Hence find the value of $p$ if $\int_{0}^{1} f (x) d x = \ln (p)$ .

[4]

Sketch the graph of $y = f (| x |)$ .

[2]

Determine the area of the region enclosed between the graph of $y = f (| x |)$ , the $x$ -axis and the lines with equations $x = - 1$ and $x = 1$ .

[3]

Markscheme

$x^{2} + 3 x + 2 = {(x + \frac{3}{2})}^{2} - \frac{1}{4}$ A1

[1 mark]

a.i.

$x^{2} + 3 x + 2 = (x + 2) (x + 1)$ A1

[1 mark]

a.ii.

M17/5/MATHL/HP1/ENG/TZ1/B11.b/M

A1 for the shape

A1 for the equation $y = 0$

A1 for asymptotes $x = - 2$ and $x = - 1$

A1 for coordinates $(- \frac{3}{2}, - 4)$

A1 $y$ -intercept $(0, \frac{1}{2})$

[5 marks]

$\int_{0}^{1} \frac{1}{x + 1} - \frac{1}{x + 2} d x$

$= {[\ln (x + 1) - \ln (x + 2)]}_{0}^{1}$ A1

$= \ln 2 - \ln 3 - \ln 1 + \ln 2$ M1

$= \ln (\frac{4}{3})$ M1A1

$∴ p = \frac{4}{3}$

[4 marks]

M17/5/MATHL/HP1/ENG/TZ1/B11.e/M

symmetry about the $y$ -axis M1

correct shape A1

Note: Allow FT from part (b).

[2 marks]

$2 \int_{0}^{1} f (x) d x$ (M1)(A1)

$= 2 \ln (\frac{4}{3})$ A1

Note: Do not award FT from part (e).

[3 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

17M.1.AHL.TZ1.H_12e.i

Question

Consider the function $q (x) = x^{5} - 10 x^{2} + 15 x - 6, x \in R$ .

Show that the graph of $y = q (x)$ is concave up for $x > 1$ .

[3]

e.i.

Sketch the graph of $y = q (x)$ showing clearly any intercepts with the axes.

[3]

e.ii.

Markscheme

$\frac{d^{2} y}{d x^{2}} = 20 x^{3} - 20$ M1A1

for $x > 1, 20 x^{3} - 20 > 0 \Rightarrow$ concave up R1AG

[3 marks]

e.i.

M17/5/MATHL/HP1/ENG/TZ1/B12.e.ii/M

$x$ -intercept at $(1, 0)$ A1

$y$ -intercept at $(0, - 6)$ A1

stationary point of inflexion at $(1, 0)$ with correct curvature either side A1

[3 marks]

e.ii.

Examiners report

[N/A]

e.i.

[N/A]

e.ii.

17N.1.AHL.TZ0.H_5

Question

A particle moves in a straight line such that at time $t$ seconds $(t ⩾ 0)$ , its velocity $v$ , in $m s^{- 1}$ , is given by $v = 10 t e^{- 2 t}$ . Find the exact distance travelled by the particle in the first half-second.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$s = \int_{0}^{\frac{1}{2}} 10 t e^{- 2 t} d t$

attempt at integration by parts M1

$= {[- 5 t e^{- 2 t}]}_{0}^{\frac{1}{2}} - \int_{0}^{\frac{1}{2}} - 5 e^{- 2 t} d t$ A1

$= {[- 5 t e^{- 2 t} - \frac{5}{2} e^{- 2 t}]}_{0}^{\frac{1}{2}}$ (A1)

Note: Condone absence of limits (or incorrect limits) and missing factor of 10 up to this point.

$s = \int_{0}^{\frac{1}{2}} 10 t e^{- 2 t} d t$ (M1)

$= - 5 e^{- 1} + \frac{5}{2} (= \frac{- 5}{e} + \frac{5}{2}) (= \frac{5 e - 10}{2 e})$ A1

[5 marks]

Examiners report

[N/A]

17N.1.AHL.TZ0.H_7

Question

The folium of Descartes is a curve defined by the equation $x^{3} + y^{3} - 3 x y = 0$ , shown in the following diagram.

N17/5/MATHL/HP1/ENG/TZ0/07

Determine the exact coordinates of the point P on the curve where the tangent line is parallel to the $y$ -axis.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$x^{3} + y^{3} - 3 x y = 0$

$3 x^{2} + 3 y^{2} \frac{d y}{d x} - 3 x \frac{d y}{d x} - 3 y = 0$ M1A1

Note: Differentiation wrt $y$ is also acceptable.

$\frac{d y}{d x} = \frac{3 y - 3 x^{2}}{3 y^{2} - 3 x} (= \frac{y - x^{2}}{y^{2} - x})$ (A1)

Note: All following marks may be awarded if the denominator is correct, but the numerator incorrect.

$y^{2} - x = 0$ M1

EITHER

$x = y^{2}$

$y^{6} + y^{3} - 3 y^{3} = 0$ M1A1

$y^{6} - 2 y^{3} = 0$

$y^{3} (y^{3} - 2) = 0$

$(y \neq 0) ∴ y = \sqrt[3]{2}$ A1

$x = {(\sqrt[3]{2})}^{2} (= \sqrt[3]{4})$ A1

$x^{3} + x y - 3 x y = 0$ M1

$x (x^{2} - 2 y) = 0$

$x \neq 0 \Rightarrow y = \frac{x^{2}}{2}$ A1

$y^{2} = \frac{x^{4}}{4}$

$x = \frac{x^{4}}{4}$

$x (x^{3} - 4) = 0$

$(x \neq 0) ∴ x = \sqrt[3]{4}$ A1

$y = \frac{{(\sqrt[3]{4})}^{2}}{2} = \sqrt[3]{2}$ A1

[8 marks]

Examiners report

[N/A]

17N.1.AHL.TZ0.H_11a

Question

Consider the function $f_{n} (x) = (\cos 2 x) (\cos 4 x) \dots (\cos 2^{n} x), n \in Z^{+}$ .

Determine whether $f_{n}$ is an odd or even function, justifying your answer.

[2]

By using mathematical induction, prove that

$f_{n} (x) = \frac{\sin 2^{n + 1} x}{2^{n} \sin 2 x}, x \neq \frac{m π}{2}$ where $m \in Z$ .

[8]

Hence or otherwise, find an expression for the derivative of $f_{n} (x)$ with respect to $x$ .

[3]

Show that, for $n > 1$ , the equation of the tangent to the curve $y = f_{n} (x)$ at $x = \frac{π}{4}$ is $4 x - 2 y - π = 0$ .

[8]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

even function A1

since $\cos k x = \cos (- k x)$ and $f_{n} (x)$ is a product of even functions R1

even function A1

since $(\cos 2 x) (\cos 4 x) \dots = (\cos (- 2 x)) (\cos (- 4 x)) \dots$ R1

Note: Do not award A0R1.

[2 marks]

consider the case $n = 1$

$\frac{\sin 4 x}{2 \sin 2 x} = \frac{2 \sin 2 x \cos 2 x}{2 \sin 2 x} = \cos 2 x$ M1

hence true for $n = 1$ R1

assume true for $n = k$ , ie, $(\cos 2 x) (\cos 4 x) \dots (\cos 2^{k} x) = \frac{\sin 2^{k + 1} x}{2^{k} \sin 2 x}$ M1

Note: Do not award M1 for “let $n = k$ ” or “assume $n = k$ ” or equivalent.

consider $n = k + 1$ :

$f_{k + 1} (x) = f_{k} (x) (\cos 2^{k + 1} x)$ (M1)

$= \frac{\sin 2^{k + 1} x}{2^{k} \sin 2 x} \cos 2^{k + 1} x$ A1

$= \frac{2 \sin 2^{k + 1} x \cos 2^{k + 1} x}{2^{k + 1} \sin 2 x}$ A1

$= \frac{\sin 2^{k + 2} x}{2^{k + 1} \sin 2 x}$ A1

so $n = 1$ true and $n = k$ true $\Rightarrow n = k + 1$ true. Hence true for all $n \in Z^{+}$ R1

Note: To obtain the final R1, all the previous M marks must have been awarded.

[8 marks]

attempt to use $f^{'} = \frac{v u^{'} - u v^{'}}{v^{2}}$ (or correct product rule) M1

$f_{n}^{'} (x) = \frac{(2^{n} \sin 2 x) (2^{n + 1} \cos 2^{n + 1} x) - (\sin 2^{n + 1} x) (2^{n + 1} \cos 2 x)}{{(2^{n} \sin 2 x)}^{2}}$ A1A1

Note: Award A1 for correct numerator and A1 for correct denominator.

[3 marks]

$f_{n}^{'} (\frac{π}{4}) = \frac{(2^{n} \sin \frac{π}{2}) (2^{n + 1} \cos 2^{n + 1} \frac{π}{4}) - (\sin 2^{n + 1} \frac{π}{4}) (2^{n + 1} \cos \frac{π}{2})}{{(2^{n} \sin \frac{π}{2})}^{2}}$ (M1)(A1)

$f_{n}^{'} (\frac{π}{4}) = \frac{(2^{n}) (2^{n + 1} \cos 2^{n + 1} \frac{π}{4})}{{(2^{n})}^{2}}$ (A1)

$= 2 \cos 2^{n + 1} \frac{π}{4} (= 2 \cos 2^{n - 1} π)$ A1

$f_{n}^{'} (\frac{π}{4}) = 2$ A1

$f_{n} (\frac{π}{4}) = 0$ A1

Note: This A mark is independent from the previous marks.

$y = 2 (x - \frac{π}{4})$ M1A1

$4 x - 2 y - π = 0$ AG

[8 marks]

Examiners report

[N/A]

18M.1.AHL.TZ1.H_4a

Question

Given that $\int_{- 2}^{2} f (x) d x = 10$ and $\int_{0}^{2} f (x) d x = 12$ , find

$\int_{- 2}^{0} (f (x) +2) d x$ .

[4]

$\int_{- 2}^{0} f (x +2) d x$ .

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\int_{- 2}^{0} f (x) d x = 10 - 12 = - 2$ (M1)(A1)

$\int_{- 2}^{0} {2 d x = [2 x]}_{- 2}^{0} = 4$ A1

$\int_{- 2}^{0} (f (x) +2) d x = 2$ A1

[4 marks]

$\int_{- 2}^{0} f (x +2) d x = \int_{0}^{2} f (x) d x$ (M1)

= 12 A1

[2 marks]

Examiners report

[N/A]

18M.1.AHL.TZ2.H_6b

Question

Consider the functions $f, g,$ defined for $x \in R$ , given by $f (x) = e^{- x} sin x$ and $g (x) = e^{- x} cos x$ .

Hence, or otherwise, find $\int_{0}^{π} e^{- x} sin x d x$ .

Markscheme

METHOD 1

Attempt to add $f^{'} (x)$ and $g^{'} (x)$ (M1)

$f^{'} (x) + g^{'} (x) = - 2 e^{- x} sin x$ A1

$\int_{0}^{π} e^{- x} sin x d x = {[- \frac{e^{- x}}{2} (sin x + cos x)]}_{0}^{π}$ (or equivalent) A1

Note: Condone absence of limits.

$= \frac{1}{2} (1 + e^{- π})$ A1

METHOD 2

$I = \int e^{- x} sin x d x$

$= - e^{- x} cos x - \int e^{- x} cos x d x$ OR $= - e^{- x} sin x + \int e^{- x} cos x d x$ M1A1

$= - e^{- x} sin x - e^{- x} cos x - \int e^{- x} sin x d x$

$I = \frac{1}{2} e^{- x} (sin x + cos x)$ A1

$\int_{0}^{π} e^{- x} sin x d x = \frac{1}{2} (1 + e^{- π})$ A1

[4 marks]

Examiners report

[N/A]

18M.1.AHL.TZ1.H_7a

Question

Let $y = arccos (\frac{x}{2})$

Find $\frac{d y}{d x}$ .

[2]

Find $\int_{0}^{1} arccos (\frac{x}{2}) d x$ .

[7]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$y = arccos (\frac{x}{2}) \Rightarrow \frac{d y}{d x} = - \frac{1}{2 \sqrt{1 - {(\frac{x}{2})}^{2}}} (= - \frac{1}{\sqrt{4 - x^{2}}})$ M1A1

Note: M1 is for use of the chain rule.

[2 marks]

attempt at integration by parts M1

$u = arccos (\frac{x}{2}) \Rightarrow \frac{d u}{d x} = - \frac{1}{\sqrt{4 - x^{2}}}$

$\frac{d v}{d x} = 1 \Rightarrow v = x$ (A1)

$\int_{0}^{1} arccos (\frac{x}{2}) d x = {[x arccos (\frac{x}{2})]}_{0}^{1} + \int_{0}^{1} \frac{1}{\sqrt{4 - x^{2}}} d x$ A1

using integration by substitution or inspection (M1)

${[x arccos (\frac{x}{2})]}_{0}^{1} + {[- {(4 - x^{2})}^{\frac{1}{2}}]}_{0}^{1}$ A1

Note: Award A1 for $- {(4 - x^{2})}^{\frac{1}{2}}$ or equivalent.

Note: Condone lack of limits to this point.

attempt to substitute limits into their integral M1

$= \frac{π}{3} - \sqrt{3} + 2$ A1

[7 marks]

Examiners report

[N/A]

18M.1.AHL.TZ2.H_8a

Question

Use the substitution $u = x^{\frac{1}{2}}$ to find $\int \frac{d x}{x^{\frac{3}{2}} + x^{\frac{1}{2}}}$ .

[4]

Hence find the value of $\frac{1}{2} \int_{1}^{9} \frac{d x}{x^{\frac{3}{2}} + x^{\frac{1}{2}}}$ , expressing your answer in the form arctan $q$ , where $q \in Q$ .

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\frac{d u}{d x} = \frac{1}{2} x^{- \frac{1}{2}}$ (accept $d u = \frac{1}{2} x^{- \frac{1}{2}} d x$ or equivalent) A1

substitution, leading to an integrand in terms of $u$ M1

$\int \frac{2 u d u}{u^{3} + u}$ or equivalent A1

= 2 arctan $(\sqrt{x}) (+ c)$ A1

[4 marks]

$\frac{1}{2} \int_{1}^{9} \frac{d x}{x^{\frac{3}{2}} + x^{\frac{1}{2}}}$ = arctan 3 − arctan 1 A1

tan(arctan 3 − arctan 1) = $\frac{3 - 1}{1 + 3 \times 1}$ (M1)

tan(arctan 3 − arctan 1) = $\frac{1}{2}$

arctan 3 − arctan 1 = arctan $\frac{1}{2}$ A1

[3 marks]

Examiners report

[N/A]

18M.1.AHL.TZ1.H_9c

Question

Let $f (x) = \frac{2 - 3 x^{5}}{2 x^{3}}, x \in R, x \neq 0$ .

The graph of $y = f (x)$ has a local maximum at A. Find the coordinates of A.

[5]

Show that there is exactly one point of inflexion, B, on the graph of $y = f (x)$ .

[5]

b.i.

The coordinates of B can be expressed in the form B $(2^{a}, b \times 2^{- 3 a})$ where a, b $\in Q$ . Find the value of a and the value of b.

[3]

b.ii.

Sketch the graph of $y = f (x)$ showing clearly the position of the points A and B.

[4]

Markscheme

attempt to differentiate (M1)

$f^{'} (x) = - 3 x^{- 4} - 3 x$ A1

Note: Award M1 for using quotient or product rule award A1 if correct derivative seen even in unsimplified form, for example $f^{'} (x) = \frac{- 15 x^{4} \times 2 x^{3} - 6 x^{2} (2 - 3 x^{5})}{{(2 x^{3})}^{2}}$ .

$- \frac{3}{x^{4}} - 3 x = 0$ M1

$\Rightarrow x^{5} = - 1 \Rightarrow x = - 1$ A1

$A (- 1, - \frac{5}{2})$ A1

[5 marks]

$f^{″} (x) = 0$ M1

$f^{″} (x) = 12 x^{- 5} - 3 (= 0)$ A1

Note: Award A1 for correct derivative seen even if not simplified.

$\Rightarrow x = \sqrt[5]{4} (= 2^{\frac{2}{5}})$ A1

hence (at most) one point of inflexion R1

Note: This mark is independent of the two A1 marks above. If they have shown or stated their equation has only one solution this mark can be awarded.

$f^{″} (x)$ changes sign at $x = \sqrt[5]{4} (= 2^{\frac{2}{5}})$ R1

so exactly one point of inflexion

[5 marks]

b.i.

$x = \sqrt[5]{4} = 2^{\frac{2}{5}} (\Rightarrow a = \frac{2}{5})$ A1

$f (2^{\frac{2}{5}}) = \frac{2 - 3 \times 2^{2}}{2 \times 2^{\frac{6}{5}}} = - 5 \times 2^{- \frac{6}{5}} (\Rightarrow b = - 5)$ (M1)A1

Note: Award M1 for the substitution of their value for $x$ into $f (x)$ .

[3 marks]

b.ii.

A1A1A1A1

A1 for shape for x < 0
A1 for shape for x > 0
A1 for maximum at A
A1 for POI at B.

Note: Only award last two A1s if A and B are placed in the correct quadrants, allowing for follow through.

[4 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

18M.1.AHL.TZ2.H_11a

Question

Show that $lo g_{r^{2}} x = \frac{1}{2} lo g_{r} x$ where $r, x \in R^{+}$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

$lo g_{r^{2}} x = \frac{lo g_{r} x}{lo g_{r} r^{2}} (= \frac{lo g_{r} x}{2 lo g_{r} r})$ M1A1

$= \frac{lo g_{r} x}{2}$ AG

[2 marks]

METHOD 2

$lo g_{r^{2}} x = \frac{1}{lo g_{x} r^{2}}$ M1

$= \frac{1}{2 lo g_{x} r}$ A1

$= \frac{lo g_{r} x}{2}$ AG

[2 marks]

Examiners report

[N/A]

18N.1.AHL.TZ0.H_7a

Question

Consider the curves $C_{1}$ and $C_{2}$ defined as follows

$C_{1} : x y = 4$ , $x > 0$

$C_{2} : y^{2} - x^{2} = 2$ , $x > 0$

Using implicit differentiation, or otherwise, find $\frac{d y}{d x}$ for each curve in terms of $x$ and $y$ .

[4]

Let P( $a$ , $b$ ) be the unique point where the curves $C_{1}$ and $C_{2}$ intersect.

Show that the tangent to $C_{1}$ at P is perpendicular to the tangent to $C_{2}$ at P.

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$C_{1} : y + x \frac{d y}{d x} = 0$ (M1)

Note: M1 is for use of both product rule and implicit differentiation.

$\Rightarrow \frac{d y}{d x} = - \frac{y}{x}$ A1

Note: Accept $- \frac{4}{x^{2}}$

$C_{2} : 2 y \frac{d y}{d x} - 2 x = 0$ (M1)

$\Rightarrow \frac{d y}{d x} = \frac{x}{y}$ A1

Note: Accept $\pm \frac{x}{\sqrt{2 + x^{2}}}$

[4 marks]

substituting $a$ and $b$ for $x$ and $y$ M1

product of gradients at P is $(- \frac{b}{a}) (\frac{a}{b}) = - 1$ or equivalent reasoning R1

Note: The R1 is dependent on the previous M1.

so tangents are perpendicular AG

[2 marks]

Examiners report

[N/A]

18N.1.AHL.TZ0.H_10a

Question

The function $f$ is defined by $f (x) = e^{x} cos^{2} x$ , where 0 ≤ $x$ ≤ 5. The curve $y = f (x)$ is shown on the following graph which has local maximum points at A and C and touches the $x$ -axis at B and D.

Use integration by parts to show that $\int e^{x} cos 2 x d x = \frac{2 e^{x}}{5} sin 2 x + \frac{e^{x}}{5} cos 2 x + c$ , $c \in R$ .

[5]

Hence, show that $\int e^{x} cos^{2} x d x = \frac{e^{x}}{5} sin 2 x + \frac{e^{x}}{10} cos 2 x + \frac{e^{x}}{2} + c$ , $c \in R$ .

[3]

Find the $x$ -coordinates of A and of C , giving your answers in the form $a + arctan b$ , where $a$ , $b \in R$ .

[6]

Find the area enclosed by the curve and the $x$ -axis between B and D, as shaded on the diagram.

[5]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

attempt at integration by parts with $u = e^{x}$ , $\frac{d v}{d x} = cos 2 x$ M1

$\int e^{x} cos 2 x d x = \frac{e^{x}}{2} sin 2 x d x - \int \frac{e^{x}}{2} sin 2 x d x$ A1

= $\frac{e^{x}}{2} sin 2 x - \frac{1}{2} (- \frac{e^{x}}{2} cos 2 x + \int \frac{e^{x}}{2} cos 2 x)$ M1A1

= $\frac{e^{x}}{2} sin 2 x + \frac{e^{x}}{4} cos 2 x - \frac{1}{4} \int e^{x} cos 2 x d x$

$∴ \frac{5}{4} \int e^{x} cos 2 x d x = \frac{e^{x}}{2} sin 2 x + \frac{e^{x}}{4} cos 2 x$ M1

$\int e^{x} cos 2 x d x = \frac{2 e^{x}}{5} sin 2 x + \frac{e^{x}}{5} cos 2 x (+ c)$ AG

METHOD 2

attempt at integration by parts with $u = cos 2 x$ , $\frac{d v}{d x} = e^{x}$ M1

$\int e^{x} cos 2 x d x = e^{x} cos 2 x + 2 \int e^{x} sin 2 x d x$ A1

$= e^{x} cos 2 x + 2 (e^{x} sin 2 x - 2 \int e^{x} cos 2 x d x)$ M1A1

$= e^{x} cos 2 x + 2 e^{x} sin 2 x - 4 \int e^{x} cos 2 x d x$

$∴ 5 \int e^{x} cos 2 x d x = e^{x} cos 2 x + 2 e^{x} sin 2 x$ M1

$\int e^{x} cos 2 x d x = \frac{2 e^{x}}{5} sin 2 x + \frac{e^{x}}{5} cos 2 x (+ c)$ AG

METHOD 3

attempt at use of table M1

A1A1

Note: A1 for first 2 lines correct, A1 for third line correct.

$\int e^{x} cos 2 x d x = e^{x} cos 2 x + 2 e^{x} sin 2 x - 4 \int e^{x} cos 2 x d x$ M1

$∴ 5 \int e^{x} cos 2 x d x = e^{x} cos 2 x + 2 e^{x} sin 2 x$ M1

$\int e^{x} cos 2 x d x = \frac{2 e^{x}}{5} sin 2 x + \frac{e^{x}}{5} cos 2 x (+ c)$ AG

[5 marks]

$\int e^{x} co s^{2} x d x = \int \frac{e^{x}}{2} (cos 2 x + 1) d x$ M1A1

$= \frac{1}{2} (\frac{2 e^{x}}{5} sin 2 x + \frac{e^{x}}{5} cos 2 x) + \frac{e^{x}}{2}$ A1

$= \frac{e^{x}}{5} sin 2 x + \frac{e^{x}}{10} cos 2 x + \frac{e^{x}}{2} (+ c)$ AG

Note: Do not accept solutions where the RHS is differentiated.

[3 marks]

$f^{'} (x) = e^{x} co s^{2} x - 2 e^{x} sin x cos x$ M1A1

Note: Award M1 for an attempt at both the product rule and the chain rule.

$e^{x} cos x (cos x - 2 sin x) = 0$ (M1)

Note: Award M1 for an attempt to factorise $cos x$ or divide by $cos x (cos x \neq 0)$ .

discount $cos x = 0$ (as this would also be a zero of the function)

$\Rightarrow cos x - 2 sin x = 0$

$\Rightarrow tan x = \frac{1}{2}$ (M1)

$\Rightarrow x = arctan (\frac{1}{2})$ (at A) and $x = π + arctan (\frac{1}{2})$ (at C) A1A1

Note: Award A1 for each correct answer. If extra values are seen award A1A0.

[6 marks]

$cos x = 0 \Rightarrow x = \frac{π}{2}$ or $\frac{3 π}{2}$ A1

Note: The A1may be awarded for work seen in part (c).

$\int_{\frac{π}{2}}^{\frac{3 π}{2}} (e^{x} co s^{2} x) d x = {[\frac{e^{x}}{5} sin 2 x + \frac{e^{x}}{10} cos 2 x + \frac{e^{x}}{2}]}_{\frac{π}{2}}^{\frac{3 π}{2}}$ M1

$= (- \frac{e^{\frac{3 π}{2}}}{10} + \frac{e^{\frac{3 π}{2}}}{2}) - (- \frac{e^{\frac{π}{2}}}{10} + \frac{e^{\frac{π}{2}}}{2}) (= \frac{2 e^{\frac{3 π}{2}}}{5} - \frac{2 e^{\frac{π}{2}}}{5})$ M1(A1)A1

Note: Award M1 for substitution of the end points and subtracting, (A1) for $sin 3 π = sin π = 0$ and $cos 3 π = cos π = - 1$ and A1 for a completely correct answer.

[5 marks]

Examiners report

[N/A]

19M.1.AHL.TZ2.H_4

Question

Using the substitution $u = sin x$ , find $\int \frac{co s^{3} x d x}{\sqrt{sin x}}$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$u = sin x \Rightarrow d u = cos x d x$ (A1)

valid attempt to write integral in terms of $u$ and $d u$ M1

$\int \frac{co s^{3} x d x}{\sqrt{sin x}} = \int \frac{(1 - u^{2}) d u}{\sqrt{u}}$ A1

$= \int (u^{- \frac{1}{2}} - u^{\frac{3}{2}}) d u$

$= 2 u^{\frac{1}{2}} - \frac{2 u^{\frac{5}{2}}}{5} (+ c)$ (A1)

$= 2 \sqrt{sin x} - \frac{2 {(\sqrt{sin x})}^{5}}{5} (+ c)$ or equivalent A1

[5 marks]

Examiners report

[N/A]

19M.1.AHL.TZ1.H_5

Question

A camera at point C is 3 m from the edge of a straight section of road as shown in the following diagram. The camera detects a car travelling along the road at $t$ = 0. It then rotates, always pointing at the car, until the car passes O, the point on the edge of the road closest to the camera.

A car travels along the road at a speed of 24 ms⁻¹. Let the position of the car be X and let OĈX = θ.

Find $\frac{d θ}{d t}$ , the rate of rotation of the camera, in radians per second, at the instant the car passes the point O .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

let OX = $x$

METHOD 1

$\frac{d x}{d t} = 24$ (or −24) (A1)

$\frac{d θ}{d t} = \frac{d x}{d t} \times \frac{d θ}{d x}$ (M1)

$3 tan θ = x$ A1

EITHER

$3 se c^{2} θ = \frac{d x}{d θ}$ A1

$\frac{d θ}{d t} = \frac{24}{3 se c^{2} θ}$

attempt to substitute for $θ = 0$ into their differential equation M1

$θ = arctan (\frac{x}{3})$

$\frac{d θ}{d x} = \frac{1}{3} \times \frac{1}{1 + \frac{x^{2}}{9}}$ A1

$\frac{d θ}{d t} = 24 \times \frac{1}{3 (1 + \frac{x^{2}}{9})}$

attempt to substitute for $x = 0$ into their differential equation M1

THEN

$\frac{d θ}{d t} = \frac{24}{3} = 8$ (rad s⁻¹) A1

Note: Accept −8 rad s⁻¹.

METHOD 2

$\frac{d x}{d t} = 24$ (or −24) (A1)

$3 tan θ = x$ A1

attempt to differentiate implicitly with respect to $t$ M1

$3 se c^{2} θ \times \frac{d θ}{d t} = \frac{d x}{d t}$ A1

$\frac{d θ}{d t} = \frac{24}{3 se c^{2} θ}$

attempt to substitute for $θ = 0$ into their differential equation M1

$\frac{d θ}{d t} = \frac{24}{3} = 8$ (rad s⁻¹) A1

Note: Accept −8 rad s⁻¹.

Note: Can be done by consideration of CX, use of Pythagoras.

METHOD 3

let the position of the car be at time $t$ be $d - 24 t$ from O (A1)

$tan θ = \frac{d - 24 t}{3} (= \frac{d}{3} - 8 t)$ M1

Note: For $tan θ = \frac{24 t}{3}$ award A0M1 and follow through.

EITHER

attempt to differentiate implicitly with respect to $t$ M1

$se c^{2} θ \frac{d θ}{d t} = - 8$ A1

attempt to substitute for $θ = 0$ into their differential equation M1

$θ = arctan (\frac{d}{3} - 8 t)$ M1

$\frac{d θ}{d t} = \frac{8}{1 + {(\frac{d}{3} - 8 t)}^{2}}$ A1

at O, $t = \frac{d}{24}$ A1

THEN

$\frac{d θ}{d t} = - 8$ A1

[6 marks]

Examiners report

[N/A]

19M.1.AHL.TZ2.H_6a

Question

The curve $C$ is given by the equation $y = x tan (\frac{π x y}{4})$ .

At the point (1, 1) , show that $\frac{d y}{d x} = \frac{2 + π}{2 - π}$ .

[5]

Hence find the equation of the normal to $C$ at the point (1, 1).

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to differentiate implicitly M1

$\frac{d y}{d x} = x se c^{2} (\frac{π x y}{4}) [(\frac{π}{4} x \frac{d y}{d x} + \frac{π}{4} y)] + tan (\frac{π x y}{4})$ A1A1

Note: Award A1 for each term.

attempt to substitute $x = 1$ , $y = 1$ into their equation for $\frac{d y}{d x}$ M1

$\frac{d y}{d x} = \frac{π}{2} \frac{d y}{d x} + \frac{π}{2} + 1$

$\frac{d y}{d x} (1 - \frac{π}{2}) = \frac{π}{2} + 1$ A1

$\frac{d y}{d x} = \frac{2 + π}{2 - π}$ AG

[5 marks]

attempt to use gradient of normal $= \frac{- 1}{\frac{d y}{d x}}$ (M1)

$= \frac{π - 2}{π + 2}$

so equation of normal is $y - 1 = \frac{π - 2}{π + 2} (x - 1)$ or $y = \frac{π - 2}{π + 2} x + \frac{4}{π + 2}$ A1

[2 marks]

Examiners report

[N/A]

19M.1.AHL.TZ1.H_7

Question

Find the coordinates of the points on the curve $y^{3} + 3 x y^{2} - x^{3} = 27$ at which $\frac{d y}{d x} = 0$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt at implicit differentiation M1

$3 y^{2} \frac{d y}{d x} + 3 y^{2} + 6 x y \frac{d y}{d x} - 3 x^{2} = 0$ A1A1

Note: Award A1 for the second & third terms, A1 for the first term, fourth term & RHS equal to zero.

substitution of $\frac{d y}{d x} = 0$ M1

$3 y^{2} - 3 x^{2} = 0$

$\Rightarrow y = \pm x$ A1

substitute either variable into original equation M1

$y = x \Rightarrow x^{3} = 9 \Rightarrow x = \sqrt[3]{9}$ (or $y^{3} = 9 \Rightarrow y = \sqrt[3]{9}$ ) A1

$y = - x \Rightarrow x^{3} = 27 \Rightarrow x = 3$ (or $y^{3} = - 27 \Rightarrow y = - 3$ ) A1

$(\sqrt[3]{9}, \sqrt[3]{9})$ , (3, −3) A1

[9 marks]

Examiners report

[N/A]

19M.1.AHL.TZ2.H_8a

Question

A right circular cone of radius $r$ is inscribed in a sphere with centre O and radius $R$ as shown in the following diagram. The perpendicular height of the cone is $h$ , X denotes the centre of its base and B a point where the cone touches the sphere.

Show that the volume of the cone may be expressed by $V = \frac{π}{3} (2 R h^{2} - h^{3})$ .

[4]

Given that there is one inscribed cone having a maximum volume, show that the volume of this cone is $\frac{32 π R^{3}}{81}$ .

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to use Pythagoras in triangle OXB M1

$\Rightarrow r^{2} = R^{2} - {(h - R)}^{2}$ A1

substitution of their $r^{2}$ into formula for volume of cone $V = \frac{π r^{2} h}{3}$ M1

$= \frac{π h}{3} (R^{2} - {(h - R)}^{2})$

$= \frac{π h}{3} (R^{2} - (h^{2} + R^{2} - 2 h R))$ A1

Note: This A mark is independent and may be seen anywhere for the correct expansion of ${(h - R)}^{2}$ .

$= \frac{π h}{3} (2 h R - h^{2})$

$= \frac{π}{3} (2 R h^{2} - h^{3})$ AG

[4 marks]

at max, $\frac{d V}{d h} = 0$ R1

$\frac{d V}{d h} = \frac{π}{3} (4 R h - 3 h^{2})$

$\Rightarrow 4 R h = 3 h^{2}$

$\Rightarrow h = \frac{4 R}{3}$ (since $h \neq 0$ ) A1

EITHER

$V_{max} = \frac{π}{3} (2 R h^{2} - h^{3})$ from part (a)

$= \frac{π}{3} (2 R {(\frac{4 R}{3})}^{2} - {(\frac{4 R}{3})}^{3})$ A1

$= \frac{π}{3} (2 R \frac{16 R^{2}}{9} - (\frac{64 R^{3}}{27}))$ A1

$r^{2} = R^{2} - {(\frac{4 R}{3} - R)}^{2}$

$r^{2} = R^{2} - \frac{R^{2}}{9} = \frac{8 R^{2}}{9}$ A1

$\Rightarrow V_{max} = \frac{π r^{2}}{3} (\frac{4 R}{3})$

$= \frac{4 π R}{9} (\frac{8 R^{2}}{9})$ A1

THEN

$= \frac{32 π R^{3}}{81}$ AG

[4 marks]

Examiners report

[N/A]

19M.1.AHL.TZ1.H_8a

Question

The graph of $y = f^{'} (x)$ , 0 ≤ $x$ ≤ 5 is shown in the following diagram. The curve intercepts the $x$ -axis at (1, 0) and (4, 0) and has a local minimum at (3, −1).

The shaded area enclosed by the curve $y = f^{'} (x)$ , the $x$ -axis and the $y$ -axis is 0.5. Given that $f (0) = 3$ ,

The area enclosed by the curve $y = f^{'} (x)$ and the $x$ -axis between $x = 1$ and $x = 4$ is 2.5 .

Write down the $x$ -coordinate of the point of inflexion on the graph of $y = f (x)$ .

[1]

find the value of $f (1)$ .

[3]

find the value of $f (4)$ .

[2]

Sketch the curve $y = f (x)$ , 0 ≤ $x$ ≤ 5 indicating clearly the coordinates of the maximum and minimum points and any intercepts with the coordinate axes.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

3 A1

[1 mark]

attempt to use definite integral of $f^{'} (x)$ (M1)

$\int_{0}^{1} f^{'} (x) d x = 0.5$

$f (1) - f (0) = 0.5$ (A1)

$f (1) = 0.5 + 3$

= 3.5 A1

[3 marks]

$\int_{1}^{4} f^{'} (x) d x = - 2.5$ (A1)

Note: (A1) is for −2.5.

$f (4) - f (1) = - 2.5$

$f (4) = 3.5 - 2.5$

= 1 A1

[2 marks]

A1A1A1

A1 for correct shape over approximately the correct domain
A1 for maximum and minimum (coordinates or horizontal lines from 3.5 and 1 are required),
A1 for $y$ -intercept at 3

[3 marks]

Examiners report

[N/A]

19M.1.AHL.TZ2.H_9a

Question

Consider the functions $f$ and $g$ defined on the domain $0 < x < 2 π$ by $f (x) = 3 cos 2 x$ and $g (x) = 4 - 11 cos x$ .

The following diagram shows the graphs of $y = f (x)$ and $y = g (x)$

Find the $x$ -coordinates of the points of intersection of the two graphs.

[6]

Find the exact area of the shaded region, giving your answer in the form $p π + q \sqrt{3}$ , where $p$ , $q \in Q$ .

[5]

At the points A and B on the diagram, the gradients of the two graphs are equal.

Determine the $y$ -coordinate of A on the graph of $g$ .

[6]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$3 cos 2 x = 4 - 11 cos x$

attempt to form a quadratic in $cos x$ M1

$3 (2 co s^{2} x - 1) = 4 - 11 cos x$ A1

$(6 co s^{2} x + 11 cos x - 7 = 0)$

valid attempt to solve their quadratic M1

$(3 cos x + 7) (2 cos x - 1) = 0$

$cos x = \frac{1}{2}$ A1

$x = \frac{π}{3}, \frac{5 π}{3}$ A1A1

Note: Ignore any “extra” solutions.

[6 marks]

consider (±) $\int_{\frac{π}{3}}^{\frac{5 π}{3}} (4 - 11 cos x - 3 cos 2 x) d x$ M1

$= (\pm) {[4 x - 11 sin x - \frac{3}{2} sin 2 x]}_{\frac{π}{3}}^{\frac{5 π}{3}}$ A1

Note: Ignore lack of or incorrect limits at this stage.

attempt to substitute their limits into their integral M1

$= \frac{20 π}{3} - 11 sin \frac{5 π}{3} - \frac{3}{2} sin \frac{10 π}{3} - (\frac{4 π}{3} - 11 sin \frac{π}{3} - \frac{3}{2} sin \frac{2 π}{3})$

$= \frac{16 π}{3} + \frac{11 \sqrt{3}}{2} + \frac{3 \sqrt{3}}{4} + \frac{11 \sqrt{3}}{2} + \frac{3 \sqrt{3}}{4}$

$= \frac{16 π}{3} + \frac{25 \sqrt{3}}{2}$ A1A1

[5 marks]

attempt to differentiate both functions and equate M1

$- 6 sin 2 x = 11 sin x$ A1

attempt to solve for $x$ M1

$11 sin x + 12 sin x cos x = 0$

$sin x (11 + 12 cos x) = 0$

$cos x = - \frac{11}{12} (or sin x = 0)$ A1

$\Rightarrow y = 4 - 11 (- \frac{11}{12})$ M1

$y = \frac{169}{12} (= 14 \frac{1}{12})$ A1

[6 marks]

Examiners report

[N/A]

19M.1.AHL.TZ1.H_9a

Question

Show that ${(sin x + cos x)}^{2} = 1 + sin 2 x$ .

[2]

Show that $sec 2 x + tan 2 x = \frac{cos x + sin x}{cos x - sin x}$ .

[4]

Hence or otherwise find $\int_{0}^{\frac{π}{6}} (sec 2 x + tan 2 x) d x$ in the form $ln (a + \sqrt{b})$ where $a$ , $b \in Z$ .

[9]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

${(sin x + cos x)}^{2} = si n^{2} x + 2 sin x cos x + co s^{2} x$ M1A1

Note: Do not award the M1 for just $si n^{2} x + co s^{2} x$ .

Note: Do not award A1 if correct expression is followed by incorrect working.

$= 1 + sin 2 x$ AG

[2 marks]

$sec 2 x + tan 2 x = \frac{1}{cos 2 x} + \frac{sin 2 x}{cos 2 x}$ M1

Note: M1 is for an attempt to change both terms into sine and cosine forms (with the same argument) or both terms into functions of $tan x$ .

$= \frac{1 + sin 2 x}{cos 2 x}$

$= \frac{{(sin x + cos x)}^{2}}{co s^{2} x - si n^{2} x}$ A1A1

Note: Award A1 for numerator, A1 for denominator.

$= \frac{{(sin x + cos x)}^{2}}{(cos x - sin x) (cos x + sin x)}$ M1

$= \frac{cos x + sin x}{cos x - sin x}$ AG

Note: Apply MS in reverse if candidates have worked from RHS to LHS.

Note: Alternative method using $tan 2 x$ and $sec 2 x$ in terms of $tan x$ .

[4 marks]

METHOD 1

$\int_{0}^{\frac{π}{6}} (\frac{cos x + sin x}{cos x - sin x}) d x$ A1

Note: Award A1 for correct expression with or without limits.

EITHER

$= {[- ln (cos x - sin x)]}_{0}^{\frac{π}{6}}$ or ${[ln (cos x - sin x)]}_{\frac{π}{6}}^{0}$ (M1)A1A1

Note: Award M1 for integration by inspection or substitution, A1 for $ln (cos x - sin x)$ , A1 for completely correct expression including limits.

$= - ln (cos \frac{π}{6} - sin \frac{π}{6}) + ln (cos 0 - sin 0)$ M1

Note: Award M1 for substitution of limits into their integral and subtraction.

$= - ln (\frac{\sqrt{3}}{2} - \frac{1}{2})$ (A1)

let $u = cos x - sin x$ M1

$\frac{d u}{d x} = - sin x - cos x = - (sin x + cos x)$

$- \int_{1}^{\frac{\sqrt{3}}{2} - \frac{1}{2}} (\frac{1}{u}) d u$ A1A1

Note: Award A1 for correct limits even if seen later, A1 for integral.

$= {[- ln u]}_{1}^{\frac{\sqrt{3}}{2} - \frac{1}{2}}$ or ${[ln u]}_{\frac{\sqrt{3}}{2} - \frac{1}{2}}^{1}$ A1

$= - ln (\frac{\sqrt{3}}{2} - \frac{1}{2}) (+ln 1)$ M1

THEN

$= ln (\frac{2}{\sqrt{3} - 1})$

Note: Award M1 for both putting the expression over a common denominator and for correct use of law of logarithms.

$= ln (1 + \sqrt{3})$ (M1)A1

METHOD 2

${[\frac{1}{2} ln (tan 2 x + sec 2 x) - \frac{1}{2} ln (cos 2 x)]}_{0}^{\frac{π}{6}}$ A1A1

$= \frac{1}{2} ln (\sqrt{3} + 2) - \frac{1}{2} ln (\frac{1}{2}) - 0$ A1A1(A1)

$= \frac{1}{2} ln (4 + 2 \sqrt{3})$ M1

$= \frac{1}{2} ln ({(1 + \sqrt{3})}^{2})$ M1A1

$= ln (1 + \sqrt{3})$ A1

[9 marks]

Examiners report

[N/A]

19N.1.AHL.TZ0.H_2

Question

Given that $\int_{0}^{ln k} e^{2 x} d x = 12$ , find the value of $k$ .

Markscheme

$\frac{1}{2} e^{2 x}$ seen (A1)

attempt at using limits in an integrated expression $({[\frac{1}{2} e^{2 x}]}_{0}^{ln k} = \frac{1}{2} e^{2 ln k} - \frac{1}{2} e^{0})$ (M1)

$= \frac{1}{2} e^{ln k^{2}} - \frac{1}{2} e^{0}$ (A1)

Setting their equation $= 12$ M1

Note: their equation must be an integrated expression with limits substituted.

$\frac{1}{2} k^{2} - \frac{1}{2} = 12$ A1

$(k^{2} = 25 \Rightarrow) k = 5$ A1

Note: Do not award final A1 for $k = \pm 5$ .

[6 marks]

Examiners report

[N/A]

19N.1.AHL.TZ0.H_7a

Question

Write $2 x - x^{2}$ in the form $a {(x - h)}^{2} + k$ , where $a, h, k \in R$ .

[2]

Hence, find the value of $\int_{\frac{1}{2}}^{\frac{3}{2}} \frac{1}{\sqrt{2 x - x^{2}}} d x$ .

[5]

Markscheme

attempt to complete the square or multiplication and equating coefficients (M1)

$2 x - x^{2} = - {(x - 1)}^{2} + 1$ A1

$a = - 1$ , $h = 1$ , $k = 1$

[2 marks]

use of their identity from part (a) $(\int_{\frac{1}{2}}^{\frac{3}{2}} \frac{1}{\sqrt{1 - {(x - 1)}^{2}}} d x)$ (M1)

$= {[arc sin (x - 1)]}_{\frac{1}{2}}^{\frac{3}{2}}$ or ${[arc sin (u)]}_{- \frac{1}{2}}^{\frac{1}{2}}$ A1

Note: Condone lack of, or incorrect limits up to this point.

$= arc sin (\frac{1}{2}) - arc sin (- \frac{1}{2})$ (M1)

$= \frac{π}{6} - (- \frac{π}{6})$ (A1)

$= \frac{π}{3}$ A1

[5 marks]

Examiners report

[N/A]

19N.1.AHL.TZ0.H_10c

Question

Consider $f (x) = \frac{2 x - 4}{x^{2} - 1}, - 1 < x < 1$ .

For the graph of $y = f (x)$ ,

Find $f^{'} (x)$ .

[2]

a.i.

Show that, if $f^{'} (x) = 0$ , then $x = 2 - \sqrt{3}$ .

[3]

a.ii.

find the coordinates of the $y$ -intercept.

[1]

b.i.

show that there are no $x$ -intercepts.

[2]

b.ii.

sketch the graph, showing clearly any asymptotic behaviour.

[2]

b.iii.

Show that $\frac{3}{x + 1} - \frac{1}{x - 1} = \frac{2 x - 4}{x^{2} - 1}$ .

[2]

The area enclosed by the graph of $y = f (x)$ and the line $y = 4$ can be expressed as $ln v$ . Find the value of $v$ .

[7]

Markscheme

attempt to use quotient rule (or equivalent) (M1)

$f^{'} (x) = \frac{(x^{2} - 1) (2) - (2 x - 4) (2 x)}{{(x^{2} - 1)}^{2}}$ A1

$= \frac{- 2 x^{2} + 8 x - 2}{{(x^{2} - 1)}^{2}}$

[2 marks]

a.i.

$f^{'} (x) = 0$

simplifying numerator (may be seen in part (i)) (M1)

$\Rightarrow x^{2} - 4 x + 1 = 0$ or equivalent quadratic equation A1

EITHER

use of quadratic formula

$\Rightarrow x = \frac{4 \pm \sqrt{12}}{2}$ A1

use of completing the square

${(x - 2)}^{2} = 3$ A1

THEN

$x = 2 - \sqrt{3}$ (since $2 + \sqrt{3}$ is outside the domain) AG

Note: Do not condone verification that $x = 2 - \sqrt{3} \Rightarrow f^{'} (x) = 0$ .

Do not award the final A1 as follow through from part (i).

[3 marks]

a.ii.

(0, 4) A1

[1 mark]

b.i.

$2 x - 4 = 0 \Rightarrow x = 2$ A1

outside the domain R1

[2 marks]

b.ii.

A1A1

award A1 for concave up curve over correct domain with one minimum point in the first quadrant
award A1 for approaching $x = \pm 1$ asymptotically

[2 marks]

b.iii.

valid attempt to combine fractions (using common denominator) M1

$\frac{3 (x - 1) - (x + 1)}{(x + 1) (x - 1)}$ A1

$= \frac{3 x - 3 - x - 1}{x^{2} - 1}$

$= \frac{2 x - 4}{x^{2} - 1}$ AG

[2 marks]

$f (x) = 4 \Rightarrow 2 x - 4 = 4 x^{2} - 4$ M1

( $x = 0$ or) $x = \frac{1}{2}$ A1

area under the curve is $\int_{0}^{\frac{1}{2}} f (x) d x$ M1

$= \int_{0}^{\frac{1}{2}} \frac{3}{x + 1} - \frac{1}{x - 1} d x$

Note: Ignore absence of, or incorrect limits up to this point.

$= {[3 ln | x + 1 | - ln | x - 1 |]}_{0}^{\frac{1}{2}}$ A1

$= 3 ln \frac{3}{2} - ln \frac{1}{2} (- 0)$

$= ln \frac{27}{4}$ A1

area is $2 - \int_{0}^{\frac{1}{2}} f (x) d x$ or $\int_{0}^{\frac{1}{2}} 4 d x - \int_{0}^{\frac{1}{2}} f (x) d x$ M1

$= 2 - ln \frac{27}{4}$

$= ln \frac{4 e^{2}}{27}$ A1

$(\Rightarrow v = \frac{4 e^{2}}{27})$

[7 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

[N/A]

20N.1.AHL.TZ0.F_1

Question

Use l’Hôpital’s rule to determine the value of

$\lim_{x \to 0} \frac{2 \sin x - \sin 2 x}{x^{3}}$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

using l’Hôpital’s rule,

$\lim_{x \to 0} \frac{2 \sin x - \sin 2 x}{x^{3}} = \lim_{x \to 0} \frac{2 \cos x - 2 \cos 2 x}{3 x^{2}}$ M1A1

$= \lim_{x \to 0} \frac{- 2 \sin x + 4 \sin 2 x}{6 x}$ (M1)A1

$= \lim_{x \to 0} \frac{- 2 \cos x + 8 \cos 2 x}{6}$ A1

$= 1$ A1

[6 marks]

Examiners report

[N/A]

20N.1.AHL.TZ0.H_2

Question

Find the equation of the tangent to the curve $y = e^{2 x} - 3 x$ at the point where $x = 0$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$(x = 0 \Rightarrow) y = 1$ (A1)

appreciate the need to find $\frac{d y}{d x}$ (M1)

$(\frac{d y}{d x} =) 2 e^{2 x} - 3$ A1

$(x = 0 \Rightarrow) \frac{d y}{d x} = - 1$ A1

$\frac{y - 1}{x - 0} = - 1 (y = 1 - x)$ A1

[5 marks]

Examiners report

[N/A]

20N.1.AHL.TZ0.H_11a

Question

Consider the curve $C$ defined by $y^{2} = \sin (x y), y \neq 0$ .

Show that $\frac{d y}{d x} = \frac{y \cos (x y)}{2 y - x \cos (x y)}$ .

[5]

Prove that, when $\frac{d y}{d x} = 0, y = \pm 1$ .

[5]

Hence find the coordinates of all points on $C$ , for $0 < x < 4 π$ , where $\frac{d y}{d x} = 0$ .

[5]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt at implicit differentiation M1

$2 y \frac{d y}{d x} = \cos (x y) ⌊x \frac{d y}{d x} + y⌋$ A1M1A1

Note: Award A1 for LHS, M1 for attempt at chain rule, A1 for RHS.

$2 y \frac{d y}{d x} = x \frac{d y}{d x} \cos (x y) + y \cos (x y)$

$2 y \frac{d y}{d x} - x \frac{d y}{d x} \cos (x y) = y \cos (x y)$

$\frac{d y}{d x} (2 y - x \cos (x y)) = y \cos (x y)$ M1

Note: Award M1 for collecting derivatives and factorising.

$\frac{d y}{d x} = \frac{y \cos (x y)}{2 y - x \cos (x y)}$ AG

[5 marks]

setting $\frac{d y}{d x} = 0$

$y \cos (x y) = 0$ (M1)

$(y \neq 0) \Rightarrow \cos (x y) = 0$ A1

$\Rightarrow \sin (x y) (= \pm \sqrt{1 - \cos^{2} (x y)} = \pm \sqrt{1 - 0}) = \pm 1$ OR $x y = (2 n + 1) \frac{π}{2} (n \in ℤ)$ OR $x y = \frac{π}{2}, \frac{3 π}{2}, \dots$ A1

Note: If they offer values for $x y$ , award A1 for at least two correct values in two different ‘quadrants’ and no incorrect values.

$y^{2} (= \sin (x y)) > 0$ R1

$\Rightarrow y^{2} = 1$ A1

$\Rightarrow y = \pm 1$ AG

[5 marks]

$y = \pm 1 \Rightarrow 1 = \sin (\pm x) \Rightarrow \sin x = \pm 1$ OR $y = \pm 1 \Rightarrow 0 = \cos (\pm x) \Rightarrow \cos x = 0$ (M1)

$(\sin x = 1 \Rightarrow) (\frac{π}{2}, 1), (\frac{5 π}{2}, 1)$ A1A1

$(\sin x = - 1 \Rightarrow) (\frac{3 π}{2}, - 1), (\frac{7 π}{2}, - 1)$ A1A1

Note: Allow ‘coordinates’ expressed as $x = \frac{π}{2}, y = 1$ for example.
Note: Each of the A marks may be awarded independently and are not dependent on (M1) being awarded.

Note: Mark only the candidate’s first two attempts for each case of $\sin x$ .

[5 marks]

Examiners report

[N/A]

20N.1.AHL.TZ0.H_12a

Question

Consider the function defined by $f (x) = \frac{k x - 5}{x - k}$ , where $x \in ℝ \ \{k\}$ and $k^{2} \neq 5$ .

Consider the case where $k = 3$ .

State the equation of the vertical asymptote on the graph of $y = f (x)$ .

[1]

State the equation of the horizontal asymptote on the graph of $y = f (x)$ .

[1]

Use an algebraic method to determine whether $f$ is a self-inverse function.

[4]

Sketch the graph of $y = f (x)$ , stating clearly the equations of any asymptotes and the coordinates of any points of intersections with the coordinate axes.

[3]

The region bounded by the $x$ -axis, the curve $y = f (x)$ , and the lines $x = 5$ and $x = 7$ is rotated through $2 π$ about the $x$ -axis. Find the volume of the solid generated, giving your answer in the form $π (a + b \ln 2)$ , where $a, b \in ℤ$ .

[6]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$x = k$ A1

[1 mark]

$y = k$ A1

[1 mark]

METHOD 1

$(f \circ f) (x) = \frac{k (\frac{k x - 5}{x - k}) - 5}{(\frac{k x - 5}{x - k}) - k}$ M1

$= \frac{k (k x - 5) - 5 (x - k)}{k x - 5 - k (x - k)}$ A1

$= \frac{k^{2} x - 5 k - 5 x + 5 k}{k x - 5 - k x + k^{2}}$

$= \frac{k^{2} x - 5 x}{k^{2} - 5}$ A1

$= \frac{x (k^{2} - 5)}{k^{2} - 5}$

$= x$

$(f \circ f) (x) = x$ , (hence $f$ is self-inverse) R1

Note: The statement $f (f (x)) = x$ could be seen anywhere in the candidate’s working to award R1.

METHOD 2

$f (x) = \frac{k x - 5}{x - k}$

$x = \frac{k y - 5}{y - k}$ M1

Note: Interchanging $x$ and $y$ can be done at any stage.

$x (y - k) = k y - 5$ A1

$x y - x k = k y - 5$

$x y - k y = x k - 5$

$y (x - k) = k x - 5$ A1

$y = f^{- 1} (x) = \frac{k x - 5}{x - k}$ (hence $f$ is self-inverse) R1

[4 marks]

attempt to draw both branches of a rectangular hyperbola M1

$x = 3$ and $y = 3$ A1

$(0, \frac{5}{3})$ and $(\frac{5}{3}, 0)$ A1

[3 marks]

METHOD 1

$volume = π \int_{5}^{7} {(\frac{3 x - 5}{x - 3})}^{2} d x$ (M1)

EITHER

attempt to express $\frac{3 x - 5}{x - 3}$ in the form $p + \frac{q}{x - 3}$ M1

$\frac{3 x - 5}{x - 3} = 3 + \frac{4}{x - 3}$ A1

attempt to expand ${(\frac{3 x - 5}{x - 3})}^{2}$ or ${(3 x - 5)}^{2}$ and divide out M1

${(\frac{3 x - 5}{x - 3})}^{2} = 9 + \frac{24 x - 56}{{(x - 3)}^{2}}$ A1

THEN

${(\frac{3 x - 5}{x - 3})}^{2} = 9 + \frac{24}{x - 3} + \frac{16}{{(x - 3)}^{2}}$ A1

$volume = π \int_{5}^{7} (9 + \frac{24}{x - 3} + \frac{16}{{(x - 3)}^{2}}) d x$

$= π {[9 x + 24 \ln (x - 3) - \frac{16}{x - 3}]}_{5}^{7}$ A1

$= π ⌊(63 + 24 \ln 4 - 4) - (45 + 24 \ln 2 - 8)⌋$

$= π (22 + 24 \ln 2)$ A1

METHOD 2

$volume = π \int_{5}^{7} {(\frac{3 x - 5}{x - 3})}^{2} d x$ (M1)

substituting $u = x - 3 \Rightarrow \frac{d u}{d x} = 1$ A1

$3 x - 5 = 3 (u + 3) - 5 = 3 u + 4$

$volume = π \int_{2}^{4} {(\frac{3 u + 4}{u})}^{2} d u$ M1

$= π \int_{2}^{4} 9 + \frac{16}{u^{2}} + \frac{24}{u} d u$ A1

$= π {[9 u - \frac{16}{u} + 24 \ln u]}_{2}^{4}$ A1

Note: Ignore absence of or incorrect limits seen up to this point.

$= π (22 + 24 \ln 2)$ A1

[6 marks]

Examiners report

[N/A]

21M.1.AHL.TZ1.8

Question

Use l’Hôpital’s rule to find $\lim_{x \to 0} (\frac{arctan 2 x}{\tan 3 x})$ .

Markscheme

attempt to differentiate numerator and denominator M1

$\lim_{x \to 0} (\frac{arctan 2 x}{\tan 3 x})$

$= \lim_{x \to 0} \frac{(\frac{2}{1 + 4 x^{2}})}{3 {sec}^{2} 3 x}$ A1A1

Note: A1 for numerator and A1 for denominator. Do not condone absence of limits.

attempt to substitute $x = 0$ (M1)

$= \frac{2}{3}$ A1

Note: Award a maximum of M1A1A0M1A1 for absence of limits.

[5 marks]

Examiners report

[N/A]

21M.1.AHL.TZ2.11a

Question

The acceleration, $a {ms}^{- 2}$ , of a particle moving in a horizontal line at time $t$ seconds, $t \geq 0$ , is given by $a = - (1 + v)$ where $v {ms}^{- 1}$ is the particle’s velocity and $v > - 1$ .

At $t = 0$ , the particle is at a fixed origin $O$ and has initial velocity $v_{0} {ms}^{- 1}$ .

Initially at $O$ , the particle moves in the positive direction until it reaches its maximum displacement from $O$ . The particle then returns to $O$ .

Let $s$ metres represent the particle’s displacement from $O$ and $s_{max}$ its maximum displacement from $O$ .

Let $v (T - k)$ represent the particle’s velocity $k$ seconds before it reaches $s_{max}$ , where

$v (T - k) = (1 + v_{0}) e^{- (T - k)} - 1$ .

Similarly, let $v (T + k)$ represent the particle’s velocity $k$ seconds after it reaches $s_{max}$ .

By solving an appropriate differential equation, show that the particle’s velocity at time $t$ is given by $v (t) = (1 + v_{0}) e^{- t} - 1$ .

[6]

Show that the time $T$ taken for the particle to reach $s_{max}$ satisfies the equation $e^{T} = 1 + v_{0}$ .

[2]

b.i.

By solving an appropriate differential equation and using the result from part (b) (i), find an expression for $s_{max}$ in terms of $v_{0}$ .

[5]

b.ii.

By using the result to part (b) (i), show that $v (T - k) = e^{k} - 1$ .

[2]

Deduce a similar expression for $v (T + k)$ in terms of $k$ .

[2]

Hence, show that $v (T - k) + v (T + k) \geq 0$ .

[3]

Markscheme

$\frac{d v}{d t} = - (1 + v)$ (A1)

$\int 1 d t = \int - \frac{1}{1 + v} d v$ (or equivalent / use of integrating factor) M1

$t = - \ln (1 + v) (+ C)$ A1

EITHER

attempt to find $C$ with initial conditions $t = 0, v = v_{0}$ M1

$C = \ln (1 + v_{0})$

$t = \ln (1 + v_{0}) - \ln (1 + v)$

$t = \ln (\frac{1 + v_{0}}{1 + v}) \Rightarrow e^{t} = \frac{1 + v_{0}}{1 + v}$ A1

$e^{t} (1 + v) = 1 + v_{0}$

$1 + v = (1 + v_{0}) e^{- t}$ A1

$v (t) = (1 + v_{0}) e^{- t} - 1$ AG

$t - C = - \ln (1 + v) \Rightarrow e^{t - C} = \frac{1}{(1 + v)}$

Attempt to find $C$ with initial conditions $t = 0, v = v_{0}$ M1

$e^{- C} = \frac{1}{(1 + v_{0})} \Rightarrow C = \ln (1 + v_{0})$

$t - \ln (1 + v_{0}) = - \ln (1 + v) \Rightarrow t = \ln (1 + v_{0}) - \ln (1 + v)$

$t = \ln (\frac{1 + v_{0}}{1 + v}) \Rightarrow e^{t} = \frac{1 + v_{0}}{1 + v}$ A1

$e^{t} (1 + v) = 1 + v_{0}$

$1 + v = (1 + v_{0}) e^{- t}$ A1

$v (t) = (1 + v_{0}) e^{- t} - 1$ AG

$t - C = - \ln (1 + v) \Rightarrow e^{- t + C} = 1 + v$ A1

$k e^{- t} - 1 = v$

Attempt to find $k$ with initial conditions $t = 0, v = v_{0}$ M1

$k = 1 + v_{0}$

$e^{- t} (1 + v_{0}) = 1 + v$ A1

$v (t) = (1 + v_{0}) e^{- t} - 1$ AG

Note: condone use of modulus within the ln function(s)

[6 marks]

recognition that when $t = T, v = 0$ M1

$(1 + v_{0}) e^{- T} - 1 = 0 \Rightarrow e^{- T} = \frac{1}{1 + v_{0}}$ A1

$e^{T} = 1 + v_{0}$ AG

Note: Award M1A0 for substituting $v_{0} = e^{T} - 1$ into $v$ and showing that $v = 0$ .

[6 marks]

b.i.

$s (t) = \int v (t) d t (= \int ((1 + v_{0}) e^{- t} - 1) d t)$ (M1)

$= - (1 + v_{0}) e^{- t} - t (+ D)$ A1

( $t = 0, s = 0$ so) $D = 1 + v_{0}$ A1

$s (t) = - (1 + v_{0}) e^{- t} - t + 1 + v_{0}$

at $s_{max}, e^{T} = 1 + v_{0} \Rightarrow T = \ln (1 + v_{0})$

Substituting into $s (t) (= - (1 + v_{0}) e^{- t} - t + 1 + v_{0})$ M1

$s_{max} = - (1 + v_{0}) (\frac{1}{1 + v_{0}}) - \ln (1 + v_{0}) + v_{0} + 1$ A1

$(s_{max} = v_{0} - \ln (1 + v_{0}))$

[5 marks]

b.ii.

METHOD 1

$v (T - k) = (1 + v_{0}) e^{- T} e^{k} - 1$ (M1)

$= (1 + v_{0}) (\frac{1}{1 + v_{0}}) e^{k} - 1$ A1

$= e^{k} - 1$ AG

METHOD 2

$v (T - k) = (1 + v_{0}) e^{- (T - k)} - 1$

$= e^{T} e^{- (T - k)} - 1$ M1

$= e^{T - T + k} - 1$ A1

$= e^{k} - 1$ AG

[2 marks]

METHOD 1

$v (T + k) = (1 + v_{0}) e^{- T} e^{- k} - 1$ (A1)

$= e^{- k} - 1$ A1

METHOD 2

$v (T + k) = (1 + v_{0}) e^{- (T + k)} - 1$ (A1)

$= e^{T} e^{- (T + k)} - 1$

$= e^{T - T - k} - 1$

$= e^{- k} - 1$ A1

[2 marks]

METHOD 1

$v (T - k) + v (T + k) = e^{k} + e^{- k} - 2$ A1

attempt to express as a square M1

$= {(e^{\frac{k}{2}} - e^{- \frac{k}{2}})}^{2} (\geq 0)$ A1

so $v (T - k) + v (T + k) \geq 0$ AG

METHOD 2

$v (T - k) + v (T + k) = e^{k} + e^{- k} - 2$ A1

Attempt to solve $\frac{d}{d k} (e^{k} + e^{- k}) = 0 (\Rightarrow k = 0)$ M1

minimum value of $2$ , (when $k = 0$ ), hence $e^{k} + e^{- k} \geq 2$ R1

so $v (T - k) + v (T + k) \geq 0$ AG

[3 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

21M.1.AHL.TZ1.12b

Question

Let $f (x) = \sqrt{1 + x}$ for $x > - 1$ .

Show that $f'' (x) = - \frac{1}{4 \sqrt{{(1 + x)}^{3}}}$ .

[3]

Use mathematical induction to prove that $f^{(n)} (x) = {(- \frac{1}{4})}^{n - 1} \frac{(2 n - 3)!}{(n - 2)!} {(1 + x)}^{\frac{1}{2} - n}$ for $n \in ℤ, n \geq 2$ .

[9]

Let $g (x) = e^{m x}, m \in ℚ$ .

Consider the function $h$ defined by $h (x) = f (x) \times g (x)$ for $x > - 1$ .

It is given that the $x^{2}$ term in the Maclaurin series for $h (x)$ has a coefficient of $\frac{7}{4}$ .

Find the possible values of $m$ .

[8]

Markscheme

attempt to use the chain rule M1

$f' (x) = \frac{1}{2} {(1 + x)}^{- \frac{1}{2}}$ A1

$f'' (x) = - \frac{1}{4} {(1 + x)}^{- \frac{3}{2}}$ A1

$= - \frac{1}{4 \sqrt{{(1 + x)}^{3}}}$ AG

Note: Award M1A0A0 for $f' (x) = \frac{1}{\sqrt{1 + x}}$ or equivalent seen

[3 marks]

let $n = 2$

$f^{''} (x) = (- \frac{1}{4 \sqrt{{(1 + x)}^{3}}} =) {(- \frac{1}{4})}^{1} \frac{1!}{0!} {(1 + x)}^{\frac{1}{2} - 2}$ R1

Note: Award R0 for not starting at $n = 2$ . Award subsequent marks as appropriate.

assume true for $n = k$ , (so $f^{(k)} (x) = {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 3)!}{(k - 2)!} {(1 + x)}^{\frac{1}{2} - k}$ ) M1

Note: Do not award M1 for statements such as “let $n = k$ ” or “ $n = k$ is true”. Subsequent marks can still be awarded.

consider $n = k + 1$

$LHS = f^{(k + 1)} (x) = \frac{d (f^{(k)} (x))}{d x}$ M1

$= {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 3)!}{(k - 2)!} (\frac{1}{2} - k) {(1 + x)}^{\frac{1}{2} - k - 1}$ (or equivalent) A1

EITHER

$RHS = f^{(k + 1)} (x) = {(- \frac{1}{4})}^{k} \frac{(2 k - 1)!}{(k - 1)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ (or equivalent) A1

$= {(- \frac{1}{4})}^{k} \frac{(2 k - 1) (2 k - 2) (2 k - 3)!}{(k - 1) (k - 2)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ A1

Note: Award A1 for $\frac{(2 k - 1)!}{(k - 1)!} = \frac{(2 k - 1) (2 k - 2) (2 k - 3)!}{(k - 1) (k - 2)!} (= \frac{2 (2 k - 1) (2 k - 3)!}{(k - 2)!})$

$= (- \frac{1}{4}) {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 1) (2 k - 2) (2 k - 3)!}{(k - 1) (k - 2)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ A1

$(= (- \frac{1}{2}) {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 1) (2 k - 3)!}{(k - 2)!} {(1 + x)}^{\frac{1}{2} - k - 1})$

Note: Award A1 for leading coefficient of $- \frac{1}{4}$ .

$= (\frac{1}{2} - k) {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 3)!}{(k - 2)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ A1

Note: The following A marks can be awarded in any order.

$= {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 3)!}{(k - 2)!} (\frac{1 - 2 k}{2}) {(1 + x)}^{\frac{1}{2} - k - 1}$

$= (- \frac{1}{2}) {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 1) (2 k - 3)!}{(k - 2)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ A1

Note: Award A1 for isolating $(2 k - 1)$ correctly.

$= (- \frac{1}{2}) {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 1)!}{(2 k - 2) (k - 2)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ A1

Note: Award A1 for multiplying top and bottom by $(k - 1)$ or $2 (k - 1)$ .

$= (- \frac{1}{4}) {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 1)!}{(k - 1) (k - 2)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ A1

Note: Award A1 for leading coefficient of $- \frac{1}{4}$ .

$= {(- \frac{1}{4})}^{k} \frac{(2 k - 1)!}{(k - 1)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ A1

$= {(- \frac{1}{4})}^{(k + 1) - 1} \frac{(2 (k + 1) - 3)!}{((k + 1) - 2)!} {(1 + x)}^{\frac{1}{2} - (k + 1)} = RHS$

THEN

since true for $n = 2$ , and true for $n = k + 1$ if true for $n = k$ , the statement is true for all, $n \in ℤ, n \geq 2$ by mathematical induction R1

Note: To obtain the final R1, at least four of the previous marks must have been awarded.

[9 marks]

METHOD 1

$h (x) = \sqrt{1 + x} e^{m x}$

using product rule to find $h' (x)$ (M1)

$h' (x) = \sqrt{1 + x} m e^{m x} + \frac{1}{2 \sqrt{1 + x}} e^{m x}$ A1

$h'' (x) = m (\sqrt{1 + x} m e^{m x} + \frac{1}{2 \sqrt{1 + x}} e^{m x}) + \frac{1}{2 \sqrt{1 + x}} m e^{m x} - \frac{1}{4 \sqrt{{(1 + x)}^{3}}} e^{m x}$ A1

substituting $x = 0$ into $h'' (x)$ M1

$h'' (0) = m^{2} + \frac{1}{2} m + \frac{1}{2} m - \frac{1}{4} (= m^{2} + m - \frac{1}{4})$ A1

$h (x) = h (0) + x h' (0) + \frac{x^{2}}{2!} h'' (0) + \dots$

equating $x^{2}$ coefficient to $\frac{7}{4}$ M1

$\frac{h'' (0)}{2!} = \frac{7}{4} (\Rightarrow h'' (0) = \frac{7}{2})$

$4 m^{2} + 4 m - 15 = 0$ A1

$(2 m + 5) (2 m - 3) = 0$

$m = - \frac{5}{2}$ or $m = \frac{3}{2}$ A1

METHOD 2

EITHER

attempt to find $f (0), f' (0), f'' (0)$ (M1)

$f (x) = {(1 + x)}^{\frac{1}{2}} f (0) = 1$

$f' (x) = \frac{1}{2} {(1 + x)}^{- \frac{1}{2}} f' (0) = \frac{1}{2}$

$f'' (x) = - \frac{1}{4} {(1 + x)}^{- \frac{3}{2}} f'' (0) = - \frac{1}{4}$

$f (x) = 1 + \frac{1}{2} x - \frac{1}{8} x^{2} + \dots$ A1

attempt to apply binomial theorem for rational exponents (M1)

$f (x) = {(1 + x)}^{\frac{1}{2}} = 1 + \frac{1}{2} x + \frac{(\frac{1}{2}) (- \frac{1}{2})}{2!} x^{2} \dots$

$f (x) = 1 + \frac{1}{2} x - \frac{1}{8} x^{2} + \dots$ A1

THEN

$g (x) = 1 + m x + \frac{m^{2}}{2} x^{2} + \dots$ (A1)

$h (x) = (1 + \frac{1}{2} x - \frac{1}{8} x^{2} + \dots) (1 + m x + \frac{m^{2}}{2} x^{2} + \dots)$ (M1)

coefficient of $x^{2}$ is $\frac{m^{2}}{2} + \frac{m}{2} - \frac{1}{8}$ A1

attempt to set equal to $\frac{7}{4}$ and solve M1

$\frac{m^{2}}{2} + \frac{m}{2} - \frac{1}{8} = \frac{7}{4}$

$4 m^{2} + 4 m - 15 = 0$ A1

$(2 m + 5) (2 m - 3) = 0$

$m = - \frac{5}{2}$ or $m = \frac{3}{2}$ A1

METHOD 3

$g' (x) = m e^{m x}$ and $g'' (x) = m^{2} e^{m x}$ (A1)

$h (x) = h (0) + x h' (0) + \frac{x^{2}}{2!} h'' (0) + \dots$

equating $x^{2}$ coefficient to $\frac{7}{4}$ M1

$\frac{h'' (0)}{2!} = \frac{7}{4} (\Rightarrow h'' (0) = \frac{7}{2})$

using product rule to find $h' (x)$ and $h'' (x)$ (M1)

$h' (x) = f (x) g' (x) + f' (x) g (x)$

$h'' (x) = f (x) g'' (x) + 2 f' (x) g' (x) + f'' (x) g (x)$ A1

substituting $x = 0$ into $h'' (x)$ M1

$h'' (0) = f (0) g'' (0) + 2 g' (0) f' (0) + g (0) f'' (0)$

$= 1 \times m^{2} + 2 m \times \frac{1}{2} + 1 \times (- \frac{1}{4}) (= m^{2} + m - \frac{1}{4})$ A1

$4 m^{2} + 4 m - 15 = 0$ A1

$(2 m + 5) (2 m - 3) = 0$

$m = - \frac{5}{2}$ or $m = \frac{3}{2}$ A1

[8 marks]

Examiners report

[N/A]

21N.1.AHL.TZ0.8

Question

Solve the differential equation $\frac{d y}{d x} = \frac{\ln 2 x}{x^{2}} - \frac{2 y}{x}, x > 0$ , given that $y = 4$ at $x = \frac{1}{2}$ .

Give your answer in the form $y = f (x)$ .

Markscheme

$\frac{d y}{d x} + \frac{2 y}{x} = \frac{\ln 2 x}{x^{2}}$ (M1)

attempt to find integrating factor (M1)

$(e^{\int \frac{2}{x} d x} = e^{2 \ln x}) = x^{2}$ (A1)

$x^{2} \frac{d y}{d x} + 2 x y = \ln 2 x$

$\frac{d}{d x} (x^{2} y) = \ln 2 x$

$x^{2} y = \int \ln 2 x d x$

attempt to use integration by parts (M1)

$x^{2} y = x \ln 2 x - x (+ c)$ A1

$y = \frac{\ln 2 x}{x} - \frac{1}{x} + \frac{c}{x^{2}}$

substituting $x = \frac{1}{2}, y = 4$ into an integrated equation involving $c$ M1

$4 = 0 - 2 + 4 c$

$\Rightarrow c = \frac{3}{2}$

$y = \frac{\ln 2 x}{x} - \frac{1}{x} + \frac{3}{2 x^{2}}$ A1

[7 marks]

Examiners report

[N/A]

21N.1.AHL.TZ0.9a

Question

Consider the expression $\frac{1}{\sqrt{1 + a x}} - \sqrt{1 - x}$ where $a \in ℚ, a \neq 0$ .

The binomial expansion of this expression, in ascending powers of $x$ , as far as the term in $x^{2}$ is $4 b x + b x^{2}$ , where $b \in ℚ$ .

Find the value of $a$ and the value of $b$ .

[6]

State the restriction which must be placed on $x$ for this expansion to be valid.

[1]

Markscheme

attempt to expand binomial with negative fractional power (M1)

$\frac{1}{\sqrt{1 + a x}} = {(1 + a x)}^{- \frac{1}{2}} = 1 - \frac{a x}{2} + \frac{3 a^{2} x^{2}}{8} + \dots$ A1

$\sqrt{1 - x} = {(1 - x)}^{\frac{1}{2}} = 1 - \frac{x}{2} - \frac{x^{2}}{8} + \dots$ A1

$\frac{1}{\sqrt{1 + a x}} - \sqrt{1 - x} = \frac{(1 - a)}{2} x + (\frac{3 a^{2} + 1}{8}) x^{2} + \dots$

attempt to equate coefficients of $x$ or $x^{2}$ (M1)

$x : \frac{1 - a}{2} = 4 b; x^{2} : \frac{3 a^{2} + 1}{8} = b$

attempt to solve simultaneously (M1)

$a = - \frac{1}{3}, b = \frac{1}{6}$ A1

[6 marks]

$|x| < 1$ A1

[1 mark]

Examiners report

[N/A]

21N.1.AHL.TZ0.11a

Question

Prove by mathematical induction that $\frac{d^{n}}{d x^{n}} (x^{2} e^{x}) = [x^{2} + 2 n x + n (n - 1)] e^{x}$ for $n \in ℤ^{+}$ .

[7]

Hence or otherwise, determine the Maclaurin series of $f (x) = x^{2} e^{x}$ in ascending powers of $x$ , up to and including the term in $x^{4}$ .

[3]

Hence or otherwise, determine the value of $\lim_{x \to 0} [\frac{{(x^{2} e^{x} - x^{2})}^{3}}{x^{9}}]$ .

[4]

Markscheme

For $n = 1$

LHS: $\frac{d}{d x} (x^{2} e^{x}) = x^{2} e^{x} + 2 x e^{x} (= e^{x} (x^{2} + 2 x))$ A1

RHS: $(x^{2} + 2 (1) x + 1 (1 - 1)) e^{x} (= e^{x} (x^{2} + 2 x))$ A1

so true for $n = 1$

now assume true for $n = k$ ; i.e. $\frac{d^{k}}{d x^{k}} (x^{2} e^{x}) = [x^{2} + 2 k x + k (k - 1)] e^{x}$ M1

Note: Do not award M1 for statements such as "let $n = k$ ". Subsequent marks can still be awarded.

attempt to differentiate the RHS M1

$\frac{d^{k + 1}}{d x^{k + 1}} (x^{2} e^{x}) = \frac{d}{d x} ([x^{2} + 2 k x + k (k - 1)] e^{x})$

$= (2 x + 2 k) e^{x} + (x^{2} + 2 k x + k (k - 1)) e^{x}$ A1

$= [x^{2} + 2 (k + 1) x + k (k + 1)] e^{x}$ A1

so true for $n = k$ implies true for $n = k + 1$

therefore $n = 1$ true and $n = k$ true $\Rightarrow n = k + 1$ true

therefore, true for all $n \in ℤ^{+}$ R1

Note: Award R1 only if three of the previous four marks have been awarded

[7 marks]

METHOD 1

attempt to use $\frac{d^{n}}{d x^{n}} (x^{2} e^{x}) = [x^{2} + 2 n x + n (n - 1)] e^{x}$ (M1)

Note: For $x = 0$ , $\frac{d^{n}}{d x^{n}} {(x^{2} e^{x})}_{x = 0} = n (n - 1)$ may be seen.

$f (0) = 0, f' (0) = 0, f'' (0) = 2, f''' (0) = 6, f^{(4)} (0) = 12$

use of $f (x) = f (0) + x f' (0) + \frac{x^{2}}{2!} f'' (0) + \frac{x^{3}}{3!} f''' (0) + \frac{x^{4}}{4!} f^{(4)} (0) + \dots$ (M1)

$\Rightarrow f (x) \approx x^{2} + x^{3} + \frac{1}{2} x^{4}$ A1

METHOD 2

' $x^{2} \times$ Maclaurin series of $e^{x}$ ' (M1)

$x^{2} (1 + x + \frac{x^{2}}{2!} + \dots)$ (A1)

$\Rightarrow f (x) \approx x^{2} + x^{3} + \frac{1}{2} x^{4}$ A1

[3 marks]

METHOD 1

attempt to substitute $x^{2} e^{x} \approx x^{2} + x^{3} + \frac{1}{2} x^{4}$ into $\frac{{(x^{2} e^{x} - x^{2})}^{3}}{x^{9}}$ M1

$\frac{{(x^{2} e^{x} - x^{2})}^{3}}{x^{9}} \approx \frac{{(x^{2} + x^{3} + \frac{1}{2} x^{4} (+ \dots) - x^{2})}^{3}}{x^{9}}$ (A1)

EITHER

$= \frac{{(x^{3} + \frac{1}{2} x^{4} + \dots)}^{3}}{x^{9}}$ A1

$= \frac{x^{9} (+ higher order terms)}{x^{9}}$

${(\frac{x^{3} + \frac{1}{2} x^{4} (+ \dots)}{x^{3}})}^{3}$ A1

${(1 + \frac{1}{2} x (+ \dots))}^{3}$

THEN

$= 1 (+ higher order terms)$

so $\lim_{x \to 0} [\frac{{(x^{2} e^{x} - x^{2})}^{3}}{x^{9}}] = 1$ A1

METHOD 2

$\lim_{x \to 0} [\frac{{(x^{2} e^{x} - x^{2})}^{3}}{x^{9}}] = \lim_{x \to 0} {(\frac{x^{2} e^{x} - x^{2}}{x^{3}})}^{3}$ M1

$= \lim_{x \to 0} {(\frac{e^{x} - 1}{x})}^{3}$ (A1)

attempt to use L'Hôpital's rule M1

$= \lim_{x \to 0} {(\frac{e^{x} - 0}{1})}^{3}$

$= {[\lim_{x \to 0} e^{x}]}^{3}$

$= 1$ A1

[4 marks]

Examiners report

[N/A]

22M.1.AHL.TZ1.1

Question

Find the value of $\int_{1}^{9} (\frac{3 \sqrt{x} - 5}{\sqrt{x}}) d x$ .

Markscheme

$\int \frac{3 \sqrt{x} - 5}{\sqrt{x}} d x = \int (3 - 5 x^{- \frac{1}{2}}) d x$ (A1)

$\int \frac{3 \sqrt{x} - 5}{\sqrt{x}} d x = 3 x - 10 x^{\frac{1}{2}} (+ c)$ A1A1

substituting limits into their integrated function and subtracting (M1)

$3 (9) - 10 {(9)}^{\frac{1}{2}} - (3 (1) - 10 {(1)}^{\frac{1}{2}})$ OR $27 - 10 \times 3 - (3 - 10)$

$= 4$ A1

[5 marks]

Examiners report

A mixed response was noted for this question. Candidates who simplified the algebraic fraction before integrating were far more successful in gaining full marks in this question. Many candidates used other valid approaches such as integration by substitution and integration by parts with varying degrees of success. A small number of candidates substituted the limits without integrating.

22M.1.AHL.TZ2.6b

Question

A function $f$ is defined by $f (x) = x \sqrt{1 - x^{2}}$ where $- 1 \leq x \leq 1$ .

The graph of $y = f (x)$ is shown below.

Show that $f$ is an odd function.

[2]

The range of $f$ is $a \leq y \leq b$ , where $a, b \in ℝ$ .

Find the value of $a$ and the value of $b$ .

[6]

Markscheme

attempts to replace $x$ with $- x$ M1

$f (- x) = - x \sqrt{1 - {(- x)}^{2}}$

$= - x \sqrt{1 - {(- x)}^{2}} (= - f (x))$ A1

Note: Award M1A1 for an attempt to calculate both $f (- x)$ and $- f (- x)$ independently, showing that they are equal.
Note: Award M1A0 for a graphical approach including evidence that either the graph is invariant after rotation by $180 °$ about the origin or the graph is invariant after a reflection in the $y$ -axis and then in the $x$ -axis (or vice versa).

so $f$ is an odd function AG

[2 marks]

attempts both product rule and chain rule differentiation to find $f' (x)$ M1

$f' (x) = x \times \frac{1}{2} \times (- 2 x) \times {(1 - x^{2})}^{- \frac{1}{2}} + {(1 - x^{2})}^{\frac{1}{2}} \times 1 (= \sqrt{1 - x^{2}} - \frac{x^{2}}{\sqrt{1 - x^{2}}})$ A1

$= \frac{1 - 2 x^{2}}{\sqrt{1 - x^{2}}}$

sets their $f' (x) = 0$ M1

$\Rightarrow x = \pm \frac{1}{\sqrt{2}}$ A1

attempts to find at least one of $f (\pm \frac{1}{\sqrt{2}})$ (M1)

Note: Award M1 for an attempt to evaluate $f (x)$ at least at one of their $f' (x) = 0$ roots.

$a = - \frac{1}{2}$ and $b = \frac{1}{2}$ A1

Note: Award A1 for $- \frac{1}{2} \leq y \leq \frac{1}{2}$ .

[6 marks]

Examiners report

[N/A]

22M.1.AHL.TZ2.7

Question

By using the substitution $u = sec x$ or otherwise, find an expression for $\int_{0}^{\frac{π}{3}} {sec}^{n} x \tan x d x$ in terms of $n$ , where $n$ is a non-zero real number.

Markscheme

METHOD 1

$u = sec x \Rightarrow d u = sec x \tan x d x$ (A1)

attempts to express the integral in terms of $u$ M1

$\int_{1}^{2} u^{n - 1} d u$ A1

$= \frac{1}{n} {[u^{n}]}_{1}^{2} (= \frac{1}{n} {[{sec}^{n} x]}_{0}^{\frac{π}{3}})$ A1

Note: Condone the absence of or incorrect limits up to this point.

$= \frac{2^{n} - 1^{n}}{n}$ M1

$= \frac{2^{n} - 1}{n}$ A1

Note: Award M1 for correct substitution of their limits for $u$ into their antiderivative for $u$ (or given limits for $x$ into their antiderivative for $x$ ).

METHOD 2

$\int {sec}^{n} x \tan x d x = \int {sec}^{n - 1} x sec x \tan x d x$ (A1)

applies integration by inspection (M1)

$= \frac{1}{n} {[{sec}^{n} x]}_{0}^{\frac{π}{3}}$ A2

Note: Award A2 if the limits are not stated.

$= \frac{1}{n} ({sec}^{n} \frac{π}{3} - {sec}^{n} 0)$ M1

Note: Award M1 for correct substitution into their antiderivative.

$= \frac{2^{n} - 1}{n}$ A1

[6 marks]

Examiners report

[N/A]

22M.1.AHL.TZ1.7a

Question

The continuous random variable $X$ has probability density function

$f (x) = \{\begin{cases} \frac{k}{\sqrt{4 - 3 x^{2}}}, & 0 \leq x \leq 1 \\ 0, & otherwise . \end{cases}$

Find the value of $k$ .

[4]

Find $E (X)$ .

[4]

Markscheme

attempt to integrate $\frac{k}{\sqrt{4 - 3 x^{2}}}$ (M1)

$= k ⌊\frac{1}{\sqrt{3}} arcsin (\frac{\sqrt{3}}{2} x)⌋$ A1

Note: Award (M1)A0 for $arcsin (\frac{\sqrt{3}}{2} x)$ .
Condone absence of $k$ up to this stage.

equating their integrand to $1$ M1

$k {[\frac{1}{\sqrt{3}} arcsin (\frac{\sqrt{3}}{2} x)]}_{0}^{1} = 1$

$k = \frac{3 \sqrt{3}}{π}$ A1

[4 marks]

$E (X) = \frac{3 \sqrt{3}}{π} \int_{0}^{1} \frac{x}{\sqrt{4 - 3 x^{2}}} d x$ A1

Note: Condone absence of limits if seen at a later stage.

EITHER

attempt to integrate by inspection (M1)

$= \frac{3 \sqrt{3}}{π} \times - \frac{1}{6} \int - 6 x {(4 - 3 x^{2})}^{- \frac{1}{2}} d x$

$= \frac{3 \sqrt{3}}{π} {[- \frac{1}{3} \sqrt{4 - 3 x^{2}}]}_{0}^{1}$ A1

Note: Condone the use of $k$ up to this stage.

for example, $u = 4 - 3 x^{2} \Rightarrow \frac{d u}{d x} = - 6 x$

Note: Other substitutions may be used. For example $u = - 3 x^{2}$ .

$= - \frac{\sqrt{3}}{2 π} \int_{4}^{1} u^{- \frac{1}{2}} d u$ M1

Note: Condone absence of limits up to this stage.

$= - \frac{\sqrt{3}}{2 π} {[2 \sqrt{u}]}_{4}^{1}$ A1

Note: Condone the use of $k$ up to this stage.

THEN

$= \frac{\sqrt{3}}{π}$ A1

Note: Award A0M1A1A0 for their $k [- \frac{1}{3} \sqrt{4 - 3 x^{2}}]$ or $k [- 2 \sqrt{u}]$ for working with incorrect or no limits.

[4 marks]

Examiners report

Most candidates who attempted part (a) knew that the integrand must be equated to 1 and only a small proportion of these managed to recognize the standard integral involved here. The effect of 3 in 3x² was missed by many resulting in very few completely correct answers for this part. Part (b) proved to be challenging for vast majority of the candidates and was poorly done in general. Stronger candidates who made good progress in part (a) were often successful in part (b) as well. Most candidates used a substitution, however many struggled to make progress using this approach. Often when using a substitution, the limits were unchanged. If the function was re-written in terms of x, this did not result in an error in the final answer.

[N/A]

22M.1.AHL.TZ2.8

Question

A continuous random variable $X$ has the probability density function

$f (x) = \{\begin{matrix} \frac{2}{(b - a) (c - a)} (x - a), & a \leq x \leq c \\ \frac{2}{(b - a) (b - c)} (b - x), & c < x \leq b \\ 0, & otherwise \end{matrix}$ .

The following diagram shows the graph of $y = f (x)$ for $a \leq x \leq b$ .

Given that $c \geq \frac{a + b}{2}$ , find an expression for the median of $X$ in terms of $a, b$ and $c$ .

Markscheme

let $m$ be the median

EITHER

attempts to find the area of the required triangle M1

base is $(m - a)$ (A1)

and height is $\frac{2}{(b - a) (c - a)} (m - a)$

area $= \frac{1}{2} (m - a) \times \frac{2}{(b - a) (c - a)} (m - a) (= \frac{{(m - a)}^{2}}{(b - a) (c - a)})$ A1

attempts to integrate the correct function M1

$\int_{a}^{m} \frac{2}{(b - a) (c - a)} (x - a) d x$

$= \frac{2}{(b - a) (c - a)} {[\frac{1}{2} {(x - a)}^{2}]}_{a}^{m}$ OR $\frac{2}{(b - a) (c - a)} {[\frac{x^{2}}{2} - a x]}_{a}^{m}$ A1A1

Note: Award A1 for correct integration and A1 for correct limits.

THEN

sets up (their) $\int_{a}^{m} \frac{2}{(b - a) (c - a)} (x - a) d x$ or area $= \frac{1}{2}$ M1

Note: Award M0A0A0M1A0A0 if candidates conclude that $m > c$ and set up their area or sum of integrals $= \frac{1}{2}$ .

$\frac{{(m - a)}^{2}}{(b - a) (c - a)} = \frac{1}{2}$

$m = a \pm \sqrt{\frac{(b - a) (c - a)}{2}}$ (A1)

as $m > a$ , rejects $m = a - \sqrt{\frac{(b - a) (c - a)}{2}}$

so $m = a + \sqrt{\frac{(b - a) (c - a)}{2}}$ A1

[6 marks]

Examiners report

[N/A]

22M.1.AHL.TZ2.11a

Question

A function $f$ is defined by $f (x) = \frac{1}{x^{2} - 2 x - 3}$ , where $x \in ℝ, x \neq - 1, x \neq 3$ .

A function $g$ is defined by $g (x) = \frac{1}{x^{2} - 2 x - 3}$ , where $x \in ℝ, x > 3$ .

The inverse of $g$ is $g^{- 1}$ .

A function $h$ is defined by $h (x) = arctan \frac{x}{2}$ , where $x \in ℝ$ .

Sketch the curve $y = f (x)$ , clearly indicating any asymptotes with their equations. State the coordinates of any local maximum or minimum points and any points of intersection with the coordinate axes.

[6]

Show that $g^{- 1} (x) = 1 + \frac{\sqrt{4 x^{2} + x}}{x}$ .

[6]

b.i.

State the domain of $g^{- 1}$ .

[1]

b.ii.

Given that $(h \circ g) (a) = \frac{π}{4}$ , find the value of $a$ .

Give your answer in the form $p + \frac{q}{2} \sqrt{r}$ , where $p, q, r \in ℤ^{+}$ .

[7]

Markscheme

$y$ -intercept $(0, - \frac{1}{3})$ A1

Note: Accept an indication of $- \frac{1}{3}$ on the $y$ -axis.

vertical asymptotes $x = - 1$ and $x = 3$ A1

horizontal asymptote $y = 0$ A1

uses a valid method to find the $x$ -coordinate of the local maximum point (M1)

Note: For example, uses the axis of symmetry or attempts to solve $f' (x) = 0$ .

local maximum point $(1, - \frac{1}{4})$ A1

Note: Award (M1)A0 for a local maximum point at $x = 1$ and coordinates not given.

three correct branches with correct asymptotic behaviour and the key features in approximately correct relative positions to each other A1

[6 marks]

$x = \frac{1}{y^{2} - 2 y - 3}$ M1

Note: Award M1 for interchanging $x$ and $y$ (this can be done at a later stage).

EITHER

attempts to complete the square M1

$y^{2} - 2 y - 3 = {(y - 1)}^{2} - 4$ A1

$x = \frac{1}{{(y - 1)}^{2} - 4}$

${(y - 1)}^{2} - 4 = \frac{1}{x} ({(y - 1)}^{2} = 4 + \frac{1}{x})$ A1

$y - 1 = \pm \sqrt{4 + \frac{1}{x}} (= \pm \sqrt{\frac{4 x + 1}{x}})$

attempts to solve $x y^{2} - 2 x y - 3 x - 1 = 0$ for $y$ M1

$y = \frac{- (- 2 x) \pm \sqrt{{(- 2 x)}^{2} + 4 x (3 x + 1)}}{2 x}$ A1

Note: Award A1 even if $-$ (in $\pm$ ) is missing

$= \frac{2 x \pm \sqrt{16 x^{2} + 4 x}}{2 x}$ A1

THEN

$= 1 \pm \frac{\sqrt{4 x^{2} + x}}{x}$ A1

$y > 3$ and hence $y = 1 - \frac{\sqrt{4 x^{2} + x}}{x}$ is rejected R1

Note: Award R1 for concluding that the expression for $y$ must have the ‘ $+$ ’ sign.
The R1 may be awarded earlier for using the condition $x > 3$ .

$y = 1 + \frac{\sqrt{4 x^{2} + x}}{x}$

$g^{- 1} (x) = 1 + \frac{\sqrt{4 x^{2} + x}}{x}$ AG

[6 marks]

b.i.

domain of $g^{- 1}$ is $x > 0$ A1

[1 mark]

b.ii.

attempts to find $(h \circ g) (a)$ (M1)

$(h \circ g) (a) = arctan (\frac{g (a)}{2}) ((h \circ g) (a) = arctan (\frac{1}{2 (a^{2} - 2 a - 3)}))$ (A1)

$arctan (\frac{g (a)}{2}) = \frac{π}{4} (arctan (\frac{1}{2 (a^{2} - 2 a - 3)}) = \frac{π}{4})$

attempts to solve for $g (a)$ M1

$\Rightarrow g (a) = 2 (\frac{1}{(a^{2} - 2 a - 3)} = 2)$

EITHER

$\Rightarrow a = g^{- 1} (2)$ A1

attempts to find their $g^{- 1} (2)$ M1

$a = 1 + \frac{\sqrt{4 {(2)}^{2} + 2}}{2}$ A1

Note: Award all available marks to this stage if $x$ is used instead of $a$ .

$\Rightarrow 2 a^{2} - 4 a - 7 = 0$ A1

attempts to solve their quadratic equation M1

$a = \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} + 4 (2) (7)}}{4} (= \frac{4 \pm \sqrt{72}}{4})$ A1

Note: Award all available marks to this stage if $x$ is used instead of $a$ .

THEN

$a = 1 + \frac{3}{2} \sqrt{2}$ (as $a > 3$ ) A1

$(p = 1, q = 3, r = 2)$

Note: Award A1 for $a = 1 + \frac{1}{2} \sqrt{18}$ $(p = 1, q = 1, r = 18)$

[7 marks]

Examiners report

Part (a) was generally well done. It was pleasing to see how often candidates presented complete sketches here. Several decided to sketch using the reciprocal function. Occasionally, candidates omitted the upper branches or forgot to calculate the y-coordinate of the maximum.

Part (b): The majority of candidates knew how to start finding the inverse, and those who attempted completing the square or using the quadratic formula to solve for y made good progress (both methods equally seen). Otherwise, they got lost in the algebra. Very few explicitly justified the rejection of the negative root.

Part (c) was well done in general, with some algebraic errors seen in occasions.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

22M.1.AHL.TZ1.12a

Question

The function $f$ is defined by $f (x) = e^{x} \sin x$ , where $x \in ℝ$ .

The function $g$ is defined by $g (x) = e^{x} \cos x$ , where $x \in ℝ$ .

Find the Maclaurin series for $f (x)$ up to and including the $x^{3}$ term.

[4]

Hence, find an approximate value for $\int_{0}^{1} e^{x^{2}} \sin (x^{2}) d x$ .

[4]

Show that $g (x)$ satisfies the equation $g'' (x) = 2 (g' (x) - g (x))$ .

[4]

c.i.

Hence, deduce that $g^{(4)} (x) = 2 (g''' (x) - g'' (x))$ .

[1]

c.ii.

Using the result from part (c), find the Maclaurin series for $g (x)$ up to and including the $x^{4}$ term.

[5]

Hence, or otherwise, determine the value of $\lim_{x \to 0} \frac{e^{x} \cos x - 1 - x}{x^{3}}$ .

[3]

Markscheme

METHOD 1

recognition of both known series (M1)

$e^{x} = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \dots$ and $\sin x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + \dots$

attempt to multiply the two series up to and including $x^{3}$ term (M1)

$e^{x} \sin x = (1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \dots) (x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + \dots)$

$= x - \frac{x^{3}}{3!} + x^{2} + \frac{x^{3}}{2!} + \dots$ (A1)

$e^{x} \sin x = x + x^{2} + \frac{1}{3} x^{3} + \dots$ A1

METHOD 2

$f (x) = e^{x} \sin x$

$f' (x) = e^{x} \cos x + e^{x} \sin x$ A1

$f'' (x) = e^{x} \cos x - e^{x} \sin x + e^{x} \sin x + e^{x} \cos x (= 2 e^{x} \cos x)$

$f''' (x) = 2 e^{x} \cos x - 2 e^{x} \sin x$

$f'' (x) = 2 e^{x} \cos x$ and $f''' (x) = 2 e^{x} (\cos x - \sin x)$ A1

substitute $x = 0$ into $f$ or its derivatives to obtain Maclaurin series (M1)

$e^{x} \sin x = 0 + \frac{x}{1!} \times 1 + \frac{x^{2}}{2!} \times 2 + \frac{x^{3}}{3!} \times 2 + \dots$

$e^{x} \sin x = x + x^{2} + \frac{1}{3} x^{3} + \dots$ A1

[4 marks]

$e^{x^{2}} \sin (x^{2}) = x^{2} + x^{4} + \frac{1}{3} x^{6} + \dots$ (A1)

substituting their expression and attempt to integrate M1

$\int_{0}^{1} e^{x^{2}} \sin (x^{2}) d x \approx \int_{0}^{1} (x^{2} + x^{4} + \frac{1}{3} x^{6}) d x$

Note: Condone absence of limits up to this stage.

$= {[\frac{x^{3}}{3} + \frac{x^{5}}{5} + \frac{x^{7}}{21}]}_{0}^{1}$ A1

$= \frac{61}{105}$ A1

[4 marks]

attempt to use product rule at least once M1

$g' (x) = e^{x} \cos x - e^{x} \sin x$ A1

$g'' (x) = e^{x} \cos x - e^{x} \sin x - e^{x} \sin x - e^{x} \cos x (= - 2 e^{x} \sin x)$ A1

EITHER

$2 (g' (x) - g (x)) = 2 (e^{x} \cos x - e^{x} \sin x - e^{x} \cos x) = - 2 e^{x} \sin x$ A1

$g'' (x) = 2 (e^{x} \cos x - e^{x} \sin x - e^{x} \cos x)$ A1

THEN

$g'' (x) = 2 (g' (x) - g (x))$ AG

Note: Accept working with each side separately to obtain $- 2 e^{x} \sin x$ .

[4 marks]

c.i.

$g''' (x) = 2 (g'' (x) - g' (x))$ A1

$g^{(4)} (x) = 2 (g''' (x) - g'' (x))$ AG

Note: Accept working with each side separately to obtain $- 4 e^{x} \cos x$ .

[1 mark]

c.ii.

attempt to substitute $x = 0$ into a derivative (M1)

$g (0) = 1, g' (0) = 1, g'' (0) = 0$ A1

$g''' (0) = - 2, g^{(4)} (0) = - 4$ (A1)

attempt to substitute into Maclaurin formula (M1)

$g (x) = 1 + x - \frac{2}{3!} x^{3} - \frac{4}{4!} x^{4} + \dots (= 1 + x - \frac{1}{3} x^{3} - \frac{1}{6} x^{4} + \dots)$ A1

Note: Do not award any marks for approaches that do not use the part (c) result.

[5 marks]

METHOD 1

$\lim_{x \to 0} \frac{e^{x} \cos x - 1 - x}{x^{3}} = \lim_{x \to 0} \frac{(1 + x - \frac{1}{3} x^{3} - \frac{1}{6} x^{4} + \dots) - 1 - x}{x^{3}}$ M1

$= \lim_{x \to 0} (- \frac{1}{3} - \frac{1}{6} x + \dots)$ (A1)

$= - \frac{1}{3}$ A1

Note: Condone the omission of $+ \dots$ in their working.

METHOD 2

$\lim_{x \to 0} \frac{e^{x} \cos x - 1 - x}{x^{3}} = \frac{0}{0}$ indeterminate form, attempt to apply l'Hôpital's rule M1

$= \lim_{x \to 0} \frac{e^{x} \cos x - e^{x} \sin x - 1}{3 x^{2}} (= \lim_{x \to 0} \frac{g' (x) - 1}{3 x^{2}})$

$= \frac{0}{0}$ , using l'Hôpital's rule again

$= \lim_{x \to 0} \frac{- 2 e^{x} \sin x}{6 x} (= \lim_{x \to 0} \frac{g'' (x)}{6 x})$

$= \frac{0}{0}$ , using l'Hôpital's rule again

$= \lim_{x \to 0} \frac{- 2 e^{x} \sin x - 2 e^{x} \cos x}{6} (= \lim_{x \to 0} \frac{g''' (x)}{6})$ A1

$= - \frac{1}{3}$ A1

[3 marks]

Examiners report

Part (a) was well answered using both methods. A number failed to see the connection between parts (a) and (b). Far too often, a candidate could not add three fractions together in part (b). There were many good responses to part (c) with candidates showing results on both sides are equal. A number of candidates failed to use the result from (c) in part (d). There were some good responses to part (e), with candidates working successfully with the series from (d) or applying l'Hôpital's rule. In particular, some responses were missing the appropriate limit notation and candidates following method 2 did not always show that the initial expression was of an indeterminate form before applying l'Hôpital's rule. Many candidates did not attempt parts (d) and (e).

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

EXN.1.AHL.TZ0.6

Question

Use l’Hôpital’s rule to determine the value of $\lim_{x \to 0} (\frac{2 x \cos (x^{2})}{5 \tan x})$ .

Markscheme

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

attempts to apply l’Hôpital’s rule on $\lim_{x \to 0} (\frac{2 x \cos (x^{2})}{5 \tan x})$ M1

$= \lim_{x \to 0} (\frac{2 \cos (x^{2}) - 4 x^{2} \sin (x^{2})}{5 \sec^{2} x})$ M1A1A1

Note: Award M1 for attempting to use product and chain rule differentiation on the numerator, A1 for a correct numerator and A1 for a correct denominator. The awarding of A1 for the denominator is independent of the M1.

$= \frac{2}{5}$ A1

[5 marks]

Examiners report

[N/A]

EXN.1.AHL.TZ0.11b

Question

A function $f$ is defined by $f (x) = \frac{3}{x^{2} + 2}, x \in ℝ$ .

The region $R$ is bounded by the curve $y = f (x)$ , the $x$ -axis and the lines $x = 0$ and $x = \sqrt{6}$ . Let $A$ be the area of $R$ .

The line $x = k$ divides $R$ into two regions of equal area.

Let $m$ be the gradient of a tangent to the curve $y = f (x)$ .

Sketch the curve $y = f (x)$ , clearly indicating any asymptotes with their equations and stating the coordinates of any points of intersection with the axes.

[4]

Show that $A = \frac{\sqrt{2} π}{2}$ .

[4]

Find the value of $k$ .

[4]

Show that $m = - \frac{6 x}{{(x^{2} + 2)}^{2}}$ .

[2]

Show that the maximum value of $m$ is $\frac{27}{32} \sqrt{\frac{2}{3}}$ .

[7]

Markscheme

a curve symmetrical about the $y$ -axis with correct concavity that has a local maximum point on the positive $y$ -axis A1

a curve clearly showing that $y \to 0$ as $x \to \pm \infty$ A1

$(0, \frac{3}{2})$ A1

horizontal asymptote $y = 0$ ( $x$ -axis) A1

[4 marks]

attempts to find $\int \frac{3}{x^{2} + 2} d x$ (M1)

$= [\frac{3}{\sqrt{2}} \arctan \frac{x}{\sqrt{2}}]$ A1

Note: Award M1A0 for obtaining $[k \arctan \frac{x}{\sqrt{2}}]$ where $k \neq \frac{3}{\sqrt{2}}$ .

Note: Condone the absence of or use of incorrect limits to this stage.

$= \frac{3}{\sqrt{2}} (\arctan \sqrt{3} - \arctan 0)$ (M1)

$= \frac{3}{\sqrt{2}} \times \frac{π}{3} (= \frac{π}{\sqrt{2}})$ A1

$A = \frac{\sqrt{2} π}{2}$ AG

[4 marks]

METHOD 1

EITHER

$\int_{0}^{k} \frac{3}{x^{2} + 2} d x = \frac{\sqrt{2} π}{4}$

$\frac{3}{\sqrt{2}} \arctan \frac{k}{\sqrt{2}} = \frac{\sqrt{2} π}{4}$ (M1)

$\int_{k}^{\sqrt{6}} \frac{3}{x^{2} + 2} d x = \frac{\sqrt{2} π}{4}$

$\frac{3}{\sqrt{2}} (\arctan \sqrt{3} - \arctan \frac{k}{\sqrt{2}}) = \frac{\sqrt{2} π}{4}$ (M1)

$\arctan \sqrt{3} - \arctan \frac{k}{\sqrt{2}} = \frac{π}{6}$

THEN

$\arctan \frac{k}{\sqrt{2}} = \frac{π}{6}$ A1

$\frac{k}{\sqrt{2}} = \tan \frac{π}{6} (= \frac{1}{\sqrt{3}})$ A1

$k = \frac{\sqrt{6}}{3} (= \sqrt{\frac{2}{3}})$ A1

METHOD 2

$\int_{0}^{k} \frac{3}{x^{2} + 2} d x = \int_{k}^{\sqrt{6}} \frac{3}{x^{2} + 2} d x$

$\frac{3}{\sqrt{2}} \arctan \frac{k}{\sqrt{2}} = \frac{3}{\sqrt{2}} (\arctan \sqrt{3} - \arctan \frac{k}{\sqrt{2}})$ (M1)

$\arctan \frac{k}{\sqrt{2}} = \frac{π}{6}$ A1

$\frac{k}{\sqrt{2}} = \tan \frac{π}{6} (= \frac{1}{\sqrt{3}})$ A1

$k = \frac{\sqrt{6}}{3} (= \sqrt{\frac{2}{3}})$ A1

[4 marks]

attempts to find $\frac{d}{d x} (\frac{3}{x^{2} + 2})$ (M1)

$= (3) (- 1) (2 x) {(x^{2} + 2)}^{- 2}$ A1

so $m = - \frac{6 x}{{(x^{2} + 2)}^{2}}$ AG

[2 marks]

attempts product rule or quotient rule differentiation M1

EITHER

$\frac{d m}{d x} = (- 6 x) (- 2) (2 x) {(x^{2} + 2)}^{- 3} + {(x^{2} + 2)}^{- 2} (- 6)$ A1

$\frac{d m}{d x} = \frac{{(x^{2} + 2)}^{2} (- 6) - (- 6 x) (2) (2 x) (x^{2} + 2)}{{(x^{2} + 2)}^{4}}$ A1

Note: Award A0 if the denominator is incorrect. Subsequent marks can be awarded.

THEN

attempts to express their $\frac{d m}{d x}$ as a rational fraction with a factorized numerator M1

$\frac{d m}{d x} = \frac{6 (x^{2} + 2) (3 x^{2} - 2)}{{(x^{2} + 2)}^{4}} (= \frac{6 (3 x^{2} - 2)}{{(x^{2} + 2)}^{3}})$

attempts to solve their $\frac{d m}{d x} = 0$ for $x$ M1

$x = \pm \sqrt{\frac{2}{3}}$ A1

from the curve, the maximum value of $m$ occurs at $x = - \sqrt{\frac{2}{3}}$ R1

(the minimum value of $m$ occurs at $x = \sqrt{\frac{2}{3}}$ )

Note: Award R1 for any equivalent valid reasoning.

maximum value of $m$ is $- \frac{6 (- \sqrt{\frac{2}{3}})}{{({(- \sqrt{\frac{2}{3}})}^{2} + 2)}^{2}}$ A1

leading to a maximum value of $\frac{27}{32} \sqrt{\frac{2}{3}}$ AG

[7 marks]

Examiners report

[N/A]

SPM.1.AHL.TZ0.12a

Question

The function $f$ is defined by $f (x) = e^{sin x}$ .

Find the first two derivatives of $f (x)$ and hence find the Maclaurin series for $f (x)$ up to and including the $x^{2}$ term.

[8]

Show that the coefficient of $x^{3}$ in the Maclaurin series for $f (x)$ is zero.

[4]

Using the Maclaurin series for $arctan x$ and $e^{3 x} - 1$ , find the Maclaurin series for $arctan (e^{3 x} - 1)$ up to and including the $x^{3}$ term.

[6]

Hence, or otherwise, find $lim_{x \to 0} \frac{f (x) - 1}{arctan (e^{3 x} - 1)}$ .

[3]

Markscheme

attempting to use the chain rule to find the first derivative M1

$f^{'} (x) = (cos x) e^{sin x}$ A1

attempting to use the product rule to find the second derivative M1

$f^{″} (x) = e^{sin x} (co s^{2} x - sin x)$ (or equivalent) A1

attempting to find $f (0)$ , $f^{'} (0)$ and $f^{″} (0)$ M1

$f (0) = 1$ ; $f^{'} (0) = (cos 0) e^{sin 0} = 1$ ; $f^{″} (0) = e^{sin 0} (co s^{2} 0 - sin 0) = 1$ A1

substitution into the Maclaurin formula $f (x) = f (0) + x f^{'} (0) + \frac{x^{2}}{2!} f^{″} (0) + \dots$ M1

so the Maclaurin series for $f (x)$ up to and including the $x^{2}$ term is $1 + x + \frac{x^{2}}{2}$ A1

[8 marks]

METHOD 1

attempting to differentiate $f^{″} (x)$ M1

$f^{‴} (x) = (cos x) e^{sin x} (co s^{2} x - sin x) - (cos x) e^{sin x} (2 sin x + 1)$ (or equivalent) A2

substituting $x = 0$ into their $f^{‴} (x)$ M1

$f^{‴} (0) = 1 (1 - 0) - 1 (0 + 1) = 0$

so the coefficient of $x^{3}$ in the Maclaurin series for $f (x)$ is zero AG

METHOD 2

substituting $sin x$ into the Maclaurin series for $e^{x}$ (M1)

$e^{sin x} = 1 + sin x + \frac{si n^{2} x}{2!} + \frac{si n^{3} x}{3!} + \dots$

substituting Maclaurin series for $sin x$ M1

$e^{sin x} = 1 + (x - \frac{x^{3}}{3!} + \dots) + \frac{{(x - \frac{x^{3}}{3!} + \dots)}^{2}}{2!} + \frac{{(x - \frac{x^{3}}{3!} + \dots)}^{3}}{3!} + \dots$ A1

coefficient of $x^{3}$ is $- \frac{1}{3!} + \frac{1}{3!} = 0$ A1

so the coefficient of $x^{3}$ in the Maclaurin series for $f (x)$ is zero AG

[4 marks]

substituting $3 x$ into the Maclaurin series for $e^{x}$ M1

$e^{3 x} = 1 + 3 x + \frac{{(3 x)}^{2}}{2!} + \frac{{(3 x)}^{3}}{3!} + \dots$ A1

substituting $(e^{3 x} - 1)$ into the Maclaurin series for $arctan x$ M1

$arctan (e^{3 x} - 1) = (e^{3 x} - 1) - \frac{{(e^{3 x} - 1)}^{3}}{3} + \frac{{(e^{3 x} - 1)}^{5}}{5} - \dots$

$= (3 x + \frac{{(3 x)}^{2}}{2!} + \frac{{(3 x)}^{3}}{3!} + \dots) - \frac{{(3 x + \frac{{(3 x)}^{2}}{2!} + \frac{{(3 x)}^{3}}{3!} + \dots)}^{3}}{3} + \dots$ A1

selecting correct terms from above M1

$= (3 x + \frac{{(3 x)}^{2}}{2!} + \frac{{(3 x)}^{3}}{3!}) - \frac{{(3 x)}^{3}}{3}$

$= 3 x + \frac{9 x^{2}}{2} - \frac{9 x^{3}}{2}$ A1

[6 marks]

METHOD 1

substitution of their series M1

$lim_{x \to 0} \frac{x + \frac{x^{2}}{2} + \dots}{3 x + \frac{9 x^{2}}{2} + \dots}$ A1

$= lim_{x \to 0} \frac{1 + \frac{x}{2} + \dots}{3 + \frac{9 x}{2} + \dots}$

$= \frac{1}{3}$ A1

METHOD 2

use of l’Hôpital’s rule M1

$lim_{x \to 0} \frac{(cos x) e^{sin x}}{\frac{3 e^{3 x}}{1 + {(e^{3 x} - 1)}^{2}}}$ (or equivalent) A1

$= \frac{1}{3}$ A1

[3 marks]

Examiners report

[N/A]

16N.2.AHL.TZ0.H_6

Question

An earth satellite moves in a path that can be described by the curve $72.5 x^{2} + 71.5 y^{2} = 1$ where $x = x (t)$ and $y = y (t)$ are in thousands of kilometres and $t$ is time in seconds.

Given that $\frac{d x}{d t} = 7.75 \times 10^{- 5}$ when $x = 3.2 \times 10^{- 3}$ , find the possible values of $\frac{d y}{d t}$ .

Give your answers in standard form.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

substituting for $x$ and attempting to solve for $y$ (or vice versa) (M1)

$y = (\pm) 0.11821 \dots$ (A1)

EITHER

$145 x + 143 y \frac{d y}{d x} = 0 (\frac{d y}{d x} = - \frac{145 x}{143 y})$ M1A1

$145 x \frac{d x}{d t} + 143 y \frac{d y}{d t} = 0$ M1A1

THEN

attempting to find $\frac{d x}{d t} (\frac{d y}{d t} = - \frac{145 (3.2 \times 10^{- 3})}{143 ((\pm) 0.11821 \dots)} \times (7.75 \times 10^{- 5}))$ (M1)

$\frac{d y}{d t} = \pm 2.13 \times 10^{- 6}$ A1

Note: Award all marks except the final A1 to candidates who do not consider ±.

METHOD 2

$y = (\pm) \sqrt{\frac{1 - 72.5 x^{2}}{71.5}}$ M1A1

$\frac{d y}{d x} = (\pm) 0.0274 \dots$ (M1)(A1)

$\frac{d y}{d t} = (\pm) 0.0274 \dots \times 7.75 \times 10^{- 5}$ (M1)

$\frac{d y}{d t} = \pm 2.13 \times 10^{- 6}$ A1

Note: Award all marks except the final A1 to candidates who do not consider ±.

[6 marks]

Examiners report

[N/A]

17M.2.AHL.TZ1.H_2a

Question

The curve $C$ is defined by equation $x y - \ln y = 1, y > 0$ .

Find $\frac{d y}{d x}$ in terms of $x$ and $y$ .

[4]

Determine the equation of the tangent to $C$ at the point $(\frac{2}{e}, e)$

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$y + x \frac{d y}{d x} - \frac{1}{y} \frac{d y}{d x} = 0$ M1A1A1

Note: Award A1 for the first two terms, A1 for the third term and the 0.

$\frac{d y}{d x} = \frac{y^{2}}{1 - x y}$ A1

Note: Accept $\frac{- y^{2}}{\ln y}$ .

Note: Accept $\frac{- y}{x - \frac{1}{y}}$ .

[4 marks]

$m_{T} = \frac{e^{2}}{1 - e \times \frac{2}{e}}$ (M1)

$m_{T} = - e^{2}$ (A1)

$y - e = - e^{2} x + 2 e$

$- e^{2} x - y + 3 e = 0$ or equivalent A1

Note: Accept $y = - 7.39 x + 8.15$ .

[3 marks]

Examiners report

[N/A]

17M.2.AHL.TZ2.H_2b

Question

Consider the curve defined by the equation $4 x^{2} + y^{2} = 7$ .

Find the volume of the solid formed when the region bounded by the curve, the $x$ -axis for $x ⩾ 0$ and the $y$ -axis for $y ⩾ 0$ is rotated through $2 π$ about the $x$ -axis.

Markscheme

Use of $V = π \int_{0}^{\frac{\sqrt{7}}{2}} y^{2} d x$

$V = π \int_{0}^{\frac{\sqrt{7}}{2}} (7 - 4 x^{2}) d x$ (M1)(A1)

Note: Condone absence of limits or incorrect limits for M mark.

Do not condone absence of or multiples of $π$ .

$= 19.4 (= \frac{7 \sqrt{7} π}{3})$ A1

[3 marks]

Examiners report

[N/A]

17M.2.AHL.TZ1.H_4a

Question

The region $A$ is enclosed by the graph of $y = 2 \arcsin (x - 1) - \frac{π}{4}$ , the $y$ -axis and the line $y = \frac{π}{4}$ .

Write down a definite integral to represent the area of $A$ .

[4]

Calculate the area of $A$ .

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

$2 \arcsin (x - 1) - \frac{π}{4} = \frac{π}{4}$ (M1)

$x = 1 + \frac{1}{\sqrt{2}} (= 1.707 \dots)$ (A1)

$\int_{0}^{1 + \frac{1}{\sqrt{2}}} \frac{π}{4} - (2 \arcsin (x - 1) - \frac{π}{4}) d x$ M1A1

Note: Award M1 for an attempt to find the difference between two functions, A1 for all correct.

METHOD 2

when $x = 0, y = \frac{- 5 π}{4} (= - 3.93)$ A1

$x = 1 + \sin (\frac{4 y + π}{8})$ M1A1

Note: Award M1 for an attempt to find the inverse function.

$\int_{\frac{- 5 π}{4}}^{\frac{π}{4}} (1 + \sin (\frac{4 y + π}{8})) d y$ A1

METHOD 3

$\int_{0}^{1.38...} (2 \arcsin (x - 1) - \frac{π}{4}) d x | + \int_{0}^{1.71...} \frac{π}{4} d x - \int_{1.38...}^{1.71...} (2 \arcsin (x - 1) - \frac{π}{4}) d x$ M1A1A1A1

Note: Award M1 for considering the area below the $x$ -axis and above the $x$ -axis and A1 for each correct integral.

[4 marks]

$area = 3.30 (squareunits)$ A2

[2 marks]

Examiners report

[N/A]

17M.2.AHL.TZ1.H_8a

Question

A water trough which is 10 metres long has a uniform cross-section in the shape of a semicircle with radius 0.5 metres. It is partly filled with water as shown in the following diagram of the cross-section. The centre of the circle is O and the angle KOL is $θ$ radians.

M17/5/MATHL/HP2/ENG/TZ1/08

The volume of water is increasing at a constant rate of $0.0008 m^{3} s^{- 1}$ .

Find an expression for the volume of water $V (m^{3})$ in the trough in terms of $θ$ .

[3]

Calculate $\frac{d θ}{d t}$ when $θ = \frac{π}{3}$ .

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

area of segment $= \frac{1}{2} \times {0.5}^{2} \times (θ - \sin θ)$ M1A1

$V = areaofsegment \times 10$

$V = \frac{5}{4} (θ - \sin θ)$ A1

[3 marks]

METHOD 1

$\frac{d V}{d t} = \frac{5}{4} (1 - \cos θ) \frac{d θ}{d t}$ M1A1

$0.0008 = \frac{5}{4} (1 - \cos \frac{π}{3}) \frac{d θ}{d t}$ (M1)

$\frac{d θ}{d t} = 0.00128 (rad s^{- 1})$ A1

METHOD 2

$\frac{d θ}{d t} = \frac{d θ}{d V} \times \frac{d V}{d t}$ (M1)

$\frac{d V}{d θ} = \frac{5}{4} (1 - \cos θ)$ A1

$\frac{d θ}{d t} = \frac{4 \times 0.0008}{5 (1 - \cos \frac{π}{3})}$ (M1)

$\frac{d θ}{d t} = 0.00128 (\frac{4}{3125}) (rad s^{- 1})$ A1

[4 marks]

Examiners report

[N/A]

17M.2.AHL.TZ1.H_11a

Question

Xavier, the parachutist, jumps out of a plane at a height of $h$ metres above the ground. After free falling for 10 seconds his parachute opens. His velocity, $v m s^{- 1}$ , $t$ seconds after jumping from the plane, can be modelled by the function

$v (t) = {\begin{array}{l} 9.8 t, & 0 ⩽ t ⩽ 10 \\ \frac{98}{\sqrt{1 + {(t - 10)}^{2}}}, & t > 10 \end{array}$

His velocity when he reaches the ground is $2.8 m s^{- 1}$ .

Find his velocity when $t = 15$ .

[2]

Calculate the vertical distance Xavier travelled in the first 10 seconds.

[2]

Determine the value of $h$ .

[5]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$v (15) = \frac{98}{\sqrt{1 + {(15 - 10)}^{2}}}$ (M1)

$v (15) = 19.2 (m s^{- 1})$ A1

[2 marks]

$\int_{0}^{10} 9.8 t d t$ (M1)

$= 490 (m)$ A1

[2 marks]

$\frac{98}{\sqrt{1 + {(t - 10)}^{2}}} = 2.8$ (M1)

$t = 44.985 \dots (s)$ A1

$h = 490 + \int_{10}^{44.9...} \frac{98}{\sqrt{1 + {(t - 10)}^{2}}} d t$ (M1)(A1)

$h = 906 (m)$ A1

[5 marks]

Examiners report

[N/A]

17M.2.AHL.TZ1.H_12a

Question

Consider $f (x) = - 1 + \ln (\sqrt{x^{2} - 1})$

The function $f$ is defined by $f (x) = - 1 + \ln (\sqrt{x^{2} - 1}), x \in D$

The function $g$ is defined by $g (x) = - 1 + \ln (\sqrt{x^{2} - 1}), x \in] 1, \infty [$ .

Find the largest possible domain $D$ for $f$ to be a function.

[2]

Sketch the graph of $y = f (x)$ showing clearly the equations of asymptotes and the coordinates of any intercepts with the axes.

[3]

Explain why $f$ is an even function.

[1]

Explain why the inverse function $f^{- 1}$ does not exist.

[1]

Find the inverse function $g^{- 1}$ and state its domain.

[4]

Find $g^{'} (x)$ .

[3]

Hence, show that there are no solutions to $g^{'} (x) = 0$ ;

[2]

g.i.

Hence, show that there are no solutions to $(g^{- 1})^{'} (x) = 0$ .

[2]

g.ii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$x^{2} - 1 > 0$ (M1)

$x < - 1$ or $x > 1$ A1

[2 marks]

M17/5/MATHL/HP2/ENG/TZ1/12.b/M

shape A1

$x = 1$ and $x = - 1$ A1

$x$ -intercepts A1

[3 marks]

EITHER

$f$ is symmetrical about the $y$ -axis R1

$f (- x) = f (x)$ R1

[1 mark]

EITHER

$f$ is not one-to-one function R1

horizontal line cuts twice R1

Note: Accept any equivalent correct statement.

[1 mark]

$x = - 1 + \ln (\sqrt{y^{2} - 1})$ M1

$e^{2 x + 2} = y^{2} - 1$ M1

$g^{- 1} (x) = \sqrt{e^{2 x + 2} + 1}, x \in R$ A1A1

[4 marks]

$g^{'} (x) = \frac{1}{\sqrt{x^{2} - 1}} \times \frac{2 x}{2 \sqrt{x^{2} - 1}}$ M1A1

$g^{'} (x) = \frac{x}{x^{2} - 1}$ A1

[3 marks]

$g^{'} (x) = \frac{x}{x^{2} - 1} = 0 \Rightarrow x = 0$ M1

which is not in the domain of $g$ (hence no solutions to $g^{'} (x) = 0$ ) R1

[2 marks]

g.i.

$(g^{- 1})^{'} (x) = \frac{e^{2 x + 2}}{\sqrt{e^{2 x + 2} + 1}}$ M1

as $e^{2 x + 2} > 0 \Rightarrow (g^{- 1})^{'} (x) > 0$ so no solutions to $(g^{- 1})^{'} (x) = 0$ R1

Note: Accept: equation $e^{2 x + 2} = 0$ has no solutions.

[2 marks]

g.ii.

Examiners report

[N/A]

g.i.

[N/A]

g.ii.

17N.2.AHL.TZ0.H_8

Question

By using the substitution $x^{2} = 2 \sec θ$ , show that $\int \frac{d x}{x \sqrt{x^{4} - 4}} = \frac{1}{4} \arccos (\frac{2}{x^{2}}) + c$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

EITHER

$x^{2} = 2 \sec θ$

$2 x \frac{d x}{d θ} = 2 \sec θ \tan θ$ M1A1

$\int \frac{d x}{x \sqrt{x^{4} - 4}}$

$= \int \frac{\sec θ \tan θ d θ}{2 \sec θ \sqrt{4 \sec^{2} θ - 4}}$ M1A1

$x = \sqrt{2} (\sec θ)^{\frac{1}{2}} (= \sqrt{2} {(\cos θ)}^{- \frac{1}{2}})$

$\frac{d x}{d θ} = \frac{\sqrt{2}}{2} (\sec θ)^{\frac{1}{2}} \tan θ (= \frac{\sqrt{2}}{2} {(\cos θ)}^{- \frac{3}{2}} \sin θ)$ M1A1

$\int \frac{d x}{x \sqrt{x^{4} - 4}}$

$= \int \frac{\sqrt{2} {(\sec θ)}^{\frac{1}{2}} \tan θ d θ}{2 \sqrt{2} {(\sec θ)}^{\frac{1}{2}} \sqrt{4 \sec^{2} θ - 4}} (= \int \frac{\sqrt{2} {(\cos θ)}^{- \frac{3}{2}} \sin θ d θ}{2 \sqrt{2} {(\cos θ)}^{- \frac{1}{2}} \sqrt{4 \sec^{2} θ - 4}})$ M1A1

THEN

$= \frac{1}{2} \int \frac{\tan θ d θ}{2 \tan θ}$ (M1)

$= \frac{1}{4} \int d θ$

$= \frac{θ}{4} + c$ A1

$x^{2} = 2 \sec θ \Rightarrow \cos θ = \frac{2}{x^{2}}$ M1

Note: This M1 may be seen anywhere, including a sketch of an appropriate triangle.

so $\frac{θ}{4} + c = \frac{1}{4} \arccos (\frac{2}{x^{2}}) + c$ AG

[7 marks]

Examiners report

[N/A]

17N.2.AHL.TZ0.H_10b

Question

Consider the function $f (x) = \frac{\sqrt{x}}{\sin x}, 0 < x < π$ .

Consider the region bounded by the curve $y = f (x)$ , the $x$ -axis and the lines $x = \frac{π}{6}, x = \frac{π}{3}$ .

Show that the $x$ -coordinate of the minimum point on the curve $y = f (x)$ satisfies the equation $\tan x = 2 x$ .

[5]

a.i.

Determine the values of $x$ for which $f (x)$ is a decreasing function.

[2]

a.ii.

Sketch the graph of $y = f (x)$ showing clearly the minimum point and any asymptotic behaviour.

[3]

Find the coordinates of the point on the graph of $f$ where the normal to the graph is parallel to the line $y = - x$ .

[4]

This region is now rotated through $2 π$ radians about the $x$ -axis. Find the volume of revolution.

[3]

Markscheme

attempt to use quotient rule or product rule M1

$f^{'} (x) = \frac{\sin x (\frac{1}{2} x^{- \frac{1}{2}}) - \sqrt{x} \cos x}{\sin^{2} x} (= \frac{1}{2 \sqrt{x} \sin x} - \frac{\sqrt{x} \cos x}{\sin^{2} x})$ A1A1

Note: Award A1 for $\frac{1}{2 \sqrt{x} \sin x}$ or equivalent and A1 for $- \frac{\sqrt{x} \cos x}{\sin^{2} x}$ or equivalent.

setting $f^{'} (x) = 0$ M1

$\frac{\sin x}{2 \sqrt{x}} - \sqrt{x} \cos x = 0$

$\frac{\sin x}{2 \sqrt{x}} = \sqrt{x} \cos x$ or equivalent A1

$\tan x = 2 x$ AG

[5 marks]

a.i.

$x = 1.17$

$0 < x ⩽ 1.17$ A1A1

Note: Award A1 for $0 < x$ and A1 for $x ⩽ 1.17$ . Accept $x < 1.17$ .

[2 marks]

a.ii.

N17/5/MATHL/HP2/ENG/TZ0/10.b/M

concave up curve over correct domain with one minimum point above the $x$ -axis. A1

approaches $x = 0$ asymptotically A1

approaches $x = π$ asymptotically A1

Note: For the final A1 an asymptote must be seen, and $π$ must be seen on the $x$ -axis or in an equation.

[3 marks]

$f^{'} (x) (= \frac{\sin x (\frac{1}{2} x^{- \frac{1}{2}}) - \sqrt{x} \cos x}{\sin^{2} x}) = 1$ (A1)

attempt to solve for $x$ (M1)

$x = 1.96$ A1

$y = f (1.96 \dots)$

$= 1.51$ A1

[4 marks]

$V = π \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{x d x}{\sin^{2} x}$ (M1)(A1)

Note: M1 is for an integral of the correct squared function (with or without limits and/or $π$ ).

$= 2.68 (= 0.852 π)$ A1

[3 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

17N.2.AHL.TZ0.H_11c

Question

Consider the function $f (x) = 2 \sin^{2} x + 7 \sin 2 x + \tan x - 9, 0 ⩽ x < \frac{π}{2}$ .

Let $u = \tan x$ .

Determine an expression for $f^{'} (x)$ in terms of $x$ .

[2]

a.i.

Sketch a graph of $y = f^{'} (x)$ for $0 ⩽ x < \frac{π}{2}$ .

[4]

a.ii.

Find the $x$ -coordinate(s) of the point(s) of inflexion of the graph of $y = f (x)$ , labelling these clearly on the graph of $y = f^{'} (x)$ .

[2]

a.iii.

Express $\sin x$ in terms of $u$ .

[2]

b.i.

Express $\sin 2 x$ in terms of $u$ .

[3]

b.ii.

Hence show that $f (x) = 0$ can be expressed as $u^{3} - 7 u^{2} + 15 u - 9 = 0$ .

[2]

b.iii.

Solve the equation $f (x) = 0$ , giving your answers in the form $\arctan k$ where $k \in Z$ .

[3]

Markscheme

$f^{'} (x) = 4 \sin x \cos x + 14 \cos 2 x + \sec^{2} x$ (or equivalent) (M1)A1

[2 marks]

a.i.

N17/5/MATHL/HP2/ENG/TZ0/11.a.ii/M A1A1A1A1

Note: Award A1 for correct behaviour at $x = 0$ , A1 for correct domain and correct behaviour for $x \to \frac{π}{2}$ , A1 for two clear intersections with $x$ -axis and minimum point, A1 for clear maximum point.

[4 marks]

a.ii.

$x = 0.0736$ A1

$x = 1.13$ A1

[2 marks]

a.iii.

attempt to write $\sin x$ in terms of $u$ only (M1)

$\sin x = \frac{u}{\sqrt{1 + u^{2}}}$ A1

[2 marks]

b.i.

$\cos x = \frac{1}{\sqrt{1 + u^{2}}}$ (A1)

attempt to use $\sin 2 x = 2 \sin x \cos x (= 2 \frac{u}{\sqrt{1 + u^{2}}} \frac{1}{\sqrt{1 + u^{2}}})$ (M1)

$\sin 2 x = \frac{2 u}{1 + u^{2}}$ A1

[3 marks]

b.ii.

$2 \sin^{2} x + 7 \sin 2 x + \tan x - 9 = 0$

$\frac{2 u^{2}}{1 + u^{2}} + \frac{14 u}{1 + u^{2}} + u - 9 (= 0)$ M1

$\frac{2 u^{2} + 14 u + u (1 + u^{2}) - 9 (1 + u^{2})}{1 + u^{2}} = 0$ (or equivalent) A1

$u^{3} - 7 u^{2} + 15 u - 9 = 0$ AG

[2 marks]

b.iii.

$u = 1$ or $u = 3$ (M1)

$x = \arctan (1)$ A1

$x = \arctan (3)$ A1

Note: Only accept answers given the required form.

[3 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

a.iii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

[N/A]

18M.2.AHL.TZ1.H_5a

Question

Given that $2 x^{3} - 3 x + 1$ can be expressed in the form $A x (x^{2} + 1) + B x + C$ , find the values of the constants $A$ , $B$ and $C$ .

[2]

Hence find $\int \frac{2 x^{3} - 3 x + 1}{x^{2} + 1} d x$ .

[5]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$2 x^{3} - 3 x + 1 = A x (x^{2} + 1) + B x + C$

$A = 2, C = 1,$ A1

$A + B = - 3 \Rightarrow B = - 5$ A1

[2 marks]

$\int \frac{2 x^{3} - 3 x + 1}{x^{2} + 1} d x = \int (2 x - \frac{5 x}{x^{2} + 1} + \frac{1}{x^{2} + 1}) d x$ M1M1

Note: Award M1 for dividing by $(x^{2} + 1)$ to get $2 x$ , M1 for separating the $5 x$ and 1.

$= x^{2} - \frac{5}{2} ln (x^{2} + 1) + arctan x (+ c)$ (M1)A1A1

Note: Award (M1)A1 for integrating $\frac{5 x}{x^{2} + 1}$ , A1 for the other two terms.

[5 marks]

Examiners report

[N/A]

18M.2.AHL.TZ2.H_7a

Question

A point P moves in a straight line with velocity $v$ ms⁻¹ given by $v (t) = e^{- t} - 8 t^{2} e^{- 2 t}$ at time t seconds, where t ≥ 0.

Determine the first time t₁ at which P has zero velocity.

[2]

Find an expression for the acceleration of P at time t.

[2]

b.i.

Find the value of the acceleration of P at time t₁.

[1]

b.ii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to solve $v (t) = 0$ for t or equivalent (M1)

t₁ = 0.441(s) A1

[2 marks]

$a (t) = \frac{d v}{d t} = - e^{- t} - 16 t e^{- 2 t} + 16 t^{2} e^{- 2 t}$ M1A1

Note: Award M1 for attempting to differentiate using the product rule.

[2 marks]

b.i.

$a (t_{1}) = - 2.28$ (ms⁻²) A1

[1 mark]

b.ii.

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

18M.2.AHL.TZ1.H_9a

Question

The following graph shows the two parts of the curve defined by the equation $x^{2} y = 5 - y^{4}$ , and the normal to the curve at the point P(2 , 1).

Show that there are exactly two points on the curve where the gradient is zero.

[7]

Find the equation of the normal to the curve at the point P.

[5]

The normal at P cuts the curve again at the point Q. Find the $x$ -coordinate of Q.

[3]

The shaded region is rotated by 2 $π$ about the $y$ -axis. Find the volume of the solid formed.

[7]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

differentiating implicitly: M1

$2 x y + x^{2} \frac{d y}{d x} = - 4 y^{3} \frac{d y}{d x}$ A1A1

Note: Award A1 for each side.

if $\frac{d y}{d x} = 0$ then either $x = 0$ or $y = 0$ M1A1

$x = 0 \Rightarrow$ two solutions for $y (y = \pm \sqrt[4]{5})$ R1

$y = 0$ not possible (as 0 ≠ 5) R1

hence exactly two points AG

Note: For a solution that only refers to the graph giving two solutions at $x = 0$ and no solutions for $y = 0$ award R1 only.

[7 marks]

at (2, 1) $4 + 4 \frac{d y}{d x} = - 4 \frac{d y}{d x}$ M1

$\frac{d y}{d x} = - \frac{1}{2}$ (A1)

gradient of normal is 2 M1

1 = 4 + c (M1)

equation of normal is $y = 2 x - 3$ A1

[5 marks]

substituting (M1)

$x^{2} (2 x - 3) = 5 - {(2 x - 3)}^{4}$ or ${(\frac{y + 3}{2})}^{2} y = 5 - y^{4}$ (A1)

$x = 0.724$ A1

[3 marks]

recognition of two volumes (M1)

volume $1 = π \int_{1}^{\sqrt[4]{5}} \frac{5 - y^{4}}{y} d y (= 101 π = 3.178 \dots)$ M1A1A1

Note: Award M1 for attempt to use $π \int x^{2} d y$ , A1 for limits, A1 for $\frac{5 - y^{4}}{y}$ Condone omission of $π$ at this stage.

volume 2

EITHER

$= \frac{1}{3} π \times 2^{2} \times 4 (= 16.75 \dots)$ (M1)(A1)

$= π \int_{- 3}^{1} {(\frac{y + 3}{2})}^{2} d y (= \frac{16 π}{3} = 16.75 \dots)$ (M1)(A1)

THEN

total volume = 19.9 A1

[7 marks]

Examiners report

[N/A]

18M.2.AHL.TZ2.H_11a

Question

A curve C is given by the implicit equation $x + y - cos (x y) = 0$ .

The curve $x y = - \frac{π}{2}$ intersects C at P and Q.

Show that $\frac{d y}{d x} = - (\frac{1 + y sin (x y)}{1 + x sin (x y)})$ .

[5]

Find the coordinates of P and Q.

[4]

b.i.

Given that the gradients of the tangents to C at P and Q are m₁ and m₂ respectively, show that m₁ × m₂ = 1.

[3]

b.ii.

Find the coordinates of the three points on C, nearest the origin, where the tangent is parallel to the line $y = - x$ .

[7]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt at implicit differentiation M1

$1 + \frac{d y}{d x} + (y + x \frac{d y}{d x}) sin (x y) = 0$ A1M1A1

Note: Award A1 for first two terms. Award M1 for an attempt at chain rule A1 for last term.

$(1 + x sin (x y)) \frac{d y}{d x} = - 1 - y sin (x y)$ A1

$\frac{d y}{d x} = - (\frac{1 + y sin (x y)}{1 + x sin (x y)})$ AG

[5 marks]

EITHER

when $x y = - \frac{π}{2}, cos x y = 0$ M1

$\Rightarrow x + y = 0$ (A1)

$x - \frac{π}{2 x} - cos (\frac{- π}{2}) = 0$ or equivalent M1

$x - \frac{π}{2 x} = 0$ (A1)

THEN

therefore $x^{2} = \frac{π}{2} (x = \pm \sqrt{\frac{π}{2}}) (x = \pm 1.25)$ A1

$P (\sqrt{\frac{π}{2}}, - \sqrt{\frac{π}{2}}), Q (- \sqrt{\frac{π}{2}}, \sqrt{\frac{π}{2}})$ or $P (1.25, - 1.25), Q (- 1.25, 1.25)$ A1

[4 marks]

b.i.

m₁= $- (\frac{1 - \sqrt{\frac{π}{2}} \times - 1}{1 + \sqrt{\frac{π}{2}} \times - 1})$ M1A1

m₂= $- (\frac{1 + \sqrt{\frac{π}{2}} \times - 1}{1 - \sqrt{\frac{π}{2}} \times - 1})$ A1

m₁m₂= 1 AG

Note: Award M1A0A0 if decimal approximations are used.
Note: No FT applies.

[3 marks]

b.ii.

equate derivative to −1 M1

$(y - x) sin (x y) = 0$ (A1)

$y = x, sin (x y) = 0$ R1

in the first case, attempt to solve $2 x = cos (x^{2})$ M1

(0.486,0.486) A1

in the second case, $sin (x y) = 0 \Rightarrow x y = 0$ and $x + y = 1$ (M1)

(0,1), (1,0) A1

[7 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

18N.2.AHL.TZ0.H_2

Question

A function $f$ satisfies the conditions $f (0) = - 4$ , $f (1) = 0$ and its second derivative is $f^{″} (x) = 15 \sqrt{x} + \frac{1}{{(x + 1)}^{2}}$ , $x$ ≥ 0.

Find $f (x)$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$f^{'} (x) = \int (15 \sqrt{x} + \frac{1}{{(x + 1)}^{2}}) d x = 10 x^{\frac{3}{2}} - \frac{1}{x + 1} (+ c)$ (M1)A1A1

Note: A1 for first term, A1 for second term. Withhold one A1 if extra terms are seen.

$f (x) = \int (10 x^{\frac{3}{2}} - \frac{1}{x + 1} + c) d x = 4 x^{\frac{5}{2}} - ln (x + 1) + c x + d$ A1

Note: Allow FT from incorrect $f^{'} (x)$ if it is of the form $f^{'} (x) = A x^{\frac{3}{2}} + \frac{B}{x + 1} + c$ .

Accept $ln | x + 1 |$ .

attempt to use at least one boundary condition in their $f (x)$ (M1)

$x = 0$ , $y = - 4$

⇒ $d = - 4$ A1

$x = 1$ , $y = 0$

⇒ $0 = 4 - ln 2 + c - 4$

⇒ $c = ln 2 (= 0.693)$ A1

$f (x) = 4 x^{\frac{5}{2}} - ln (x + 1) + x ln 2 - 4$

[7 marks]

Examiners report

[N/A]

18N.2.AHL.TZ0.H_5

Question

Differentiate from first principles the function $f (x) = 3 x^{3} - x$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

$\frac{f (x + h) - f (x)}{h}$

$= \frac{(3 {(x + h)}^{3} - (x + h)) - (3 x^{3} - x)}{h}$ M1

$= \frac{3 (x^{3} + 3 x^{2} h + 3 x h^{2} + h^{3}) - x - h - 3 x^{3} + x}{h}$ (A1)

$= \frac{9 x^{2} h + 9 x h^{2} + 3 h^{3} - h}{h}$ A1

cancelling $h$ M1

$= 9 x^{2} + 9 x h + 3 h^{2} - 1$

then $lim_{h \to 0} (9 x^{2} + 9 x h + 3 h^{2} - 1)$

$= 9 x^{2} - 1$ A1

Note: Final A1 dependent on all previous marks.

METHOD 2

$\frac{f (x + h) - f (x)}{h}$

$= \frac{(3 {(x + h)}^{3} - (x + h)) - (3 x^{3} - x)}{h}$ M1

$= \frac{3 ({(x + h)}^{3} - x^{3}) + (x - (x + h))}{h}$ (A1)

$= \frac{3 h ({(x + h)}^{2} + x (x + h) + x^{2}) - h}{h}$ A1

cancelling $h$ M1

$= 3 ({(x + h)}^{2} + x (x + h) + x^{2}) - 1$

then $lim_{h \to 0} (3 ({(x + h)}^{2} + x (x + h) + x^{2}) - 1)$

$= 9 x^{2} - 1$ A1

Note: Final A1 dependent on all previous marks.

[5 marks]

Examiners report

[N/A]

18N.2.AHL.TZ0.H_9b

Question

The function $f$ is defined by $f (x) = \frac{2 ln x + 1}{x - 3}$ , 0 < $x$ < 3.

Draw a set of axes showing $x$ and $y$ values between −3 and 3. On these axes

Hence, or otherwise, find the coordinates of the point of inflexion on the graph of $y = f (x)$ .

[4]

sketch the graph of $y = f (x)$ , showing clearly any axis intercepts and giving the equations of any asymptotes.

[4]

c.i.

sketch the graph of $y = f^{- 1} (x)$ , showing clearly any axis intercepts and giving the equations of any asymptotes.

[4]

c.ii.

Hence, or otherwise, solve the inequality $f (x) > f^{- 1} (x)$ .

[3]

Markscheme

finding turning point of $y = f^{'} (x)$ or finding root of $y = f^{″} (x)$ (M1)

$x = 0.899$ A1

$y = f (0.899048 \dots) = - 0.375$ (M1)A1

(0.899, −0.375)

Note: Do not accept $x = 0.9$ . Accept y-coordinates rounding to −0.37 or −0.375 but not −0.38.

[4 marks]

smooth curve over the correct domain which does not cross the y-axis

and is concave down for $x$ > 1 A1

$x$ -intercept at 0.607 A1

equations of asymptotes given as $x$ = 0 and $x$ = 3 (the latter must be drawn) A1A1

[4 marks]

c.i.

attempt to reflect graph of $f$ in $y$ = $x$ (M1)

smooth curve over the correct domain which does not cross the $x$ -axis and is concave down for $y$ > 1 A1

$y$ -intercept at 0.607 A1

equations of asymptotes given as $y$ = 0 and $y$ = 3 (the latter must be drawn) A1

Note: For FT from (i) to (ii) award max M1A0A1A0.

[4 marks]

c.ii.

solve $f (x) = f^{- 1} (x)$ or $f (x) = x$ to get $x$ = 0.372 (M1)A1

0 < $x$ < 0.372 A1

Note: Do not award FT marks.

[3 marks]

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

19M.2.AHL.TZ1.H_1

Question

Let $l$ be the tangent to the curve $y = x e^{2 x}$ at the point (1, $e^{2}$ ).

Find the coordinates of the point where $l$ meets the $x$ -axis.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

equation of tangent is $y = 22.167 \dots x - 14.778 \dots$ OR $y = - 7.389 \dots = 22.167 \dots (x - 1)$ (M1)(A1)

meets the $x$ -axis when $y = 0$

$x = 0.667$

meets $x$ -axis at (0.667, 0) $(= (\frac{2}{3}, 0))$ A1A1

Note: Award A1 for $x = \frac{2}{3}$ or $x = 0.667$ seen and A1 for coordinates ( $x$ , 0) given.

METHOD 1

Attempt to differentiate (M1)

$\frac{d y}{d x} = e^{2 x} + 2 x e^{2 x}$

when $x = 1$ , $\frac{d y}{d x} = 3 e^{2}$ (M1)

equation of the tangent is $y - e^{2} = 3 e^{2} (x - 1)$

$y = 3 e^{2} x - 2 e^{2}$

meets $x$ -axis at $x = \frac{2}{3}$

$(\frac{2}{3}, 0)$ A1A1

Note: Award A1 for $x = \frac{2}{3}$ or $x = 0.667$ seen and A1 for coordinates ( $x$ , 0) given.

[4 marks]

Examiners report

[N/A]

19M.2.AHL.TZ1.H_4a

Question

The function $f$ is defined by $f (x) = sec x + 2$ , $0 ⩽ x < \frac{π}{2}$ .

Write down the range of $f$ .

[1]

Find $f^{- 1} (x)$ , stating its domain.

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$f (x)$ ≥ 3 A1

[1 mark]

$x = sec y + 2$ (M1)

Note: Exchange of variables can take place at any point.

$cos y = \frac{1}{x - 2}$ (A1)

$f^{'} (x) = arccos (\frac{1}{x - 2})$ , $x$ ≥ 3 A1A1

Note: Allow follow through from (a) for last A1 mark which is independent of earlier marks in (b).

[4 marks]

Examiners report

[N/A]

19M.2.AHL.TZ1.H_5

Question

The function $f$ is defined by $f (x) = sec x + 2$ , $0 ⩽ x < \frac{π}{2}$ .

Use integration by parts to find $\int {(ln x)}^{2} d x$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

write as $\int 1 \times {(ln x)}^{2} d x$ (M1)

$= x {(ln x)}^{2} - \int x \times \frac{2 (ln x)}{x} d x (= x {(ln x)}^{2} - \int 2 ln x)$ M1A1

$= x {(ln x)}^{2} - 2 x ln x + \int 2 d x$ (M1)(A1)

$= x {(ln x)}^{2} - 2 x ln x + 2 x + c$ A1

METHOD 2

let $u = ln x$ M1

$\frac{d u}{d x} = \frac{1}{x}$

$\int u^{2} e^{u} d u$ A1

$= u^{2} e^{u} - \int 2 u e^{u} d u$ M1

$= u^{2} e^{u} - 2 u e^{u} + \int 2 e^{u} d u$ A1

$= u^{2} e^{u} - 2 u e^{u} + 2 e^{u} + c$

$= x {(ln x)}^{2} - 2 x ln x + 2 x + c$ M1A1

METHOD 3

Setting up $u = ln x$ and $\frac{d v}{d x} = ln x$ M1

$ln x (x ln x - x) - \int (ln x - 1) d x$ M1A1

$= x {(ln x)}^{2} - x ln x - (x ln x - x) + x + c$ M1A1

$= x {(ln x)}^{2} - 2 x ln x + 2 x + c$ A1

[6 marks]

Examiners report

[N/A]

19M.2.AHL.TZ2.H_6

Question

A particle moves along a horizontal line such that at time $t$ seconds, $t$ ≥ 0, its acceleration $a$ is given by $a$ = 2 $t$ − 1. When $t$ = 6 , its displacement $s$ from a fixed origin O is 18.25 m. When $t$ = 15, its displacement from O is 922.75 m. Find an expression for $s$ in terms of $t$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to integrate $a$ to find $v$ M1

$v = \int a d t = \int (2 t - 1) d t$

$= t^{2} - t + c$ A1

$s = \int v d t = \int (t^{2} - t + c) d t$

$= \frac{t^{3}}{3} - \frac{t^{2}}{2} + c t + d$ A1

attempt at substitution of given values (M1)

at $t = 6, 18.25 = 72 - 18 + 6 c + d$

at $t = 15, 922.75 = 1125 - 112.5 + 15 c + d$

solve simultaneously: (M1)

$c = - 6, d = 0.25$ A1

$\Rightarrow s = \frac{t^{3}}{3} - \frac{t^{2}}{2} + - 6 t + \frac{1}{4}$

[6 marks]

Examiners report

[N/A]

19M.2.AHL.TZ1.H_7

Question

The function $f$ is defined by $f (x) = {(x - 1)}^{2}$ , $x$ ≥ 1 and the function $g$ is defined by $g (x) = x^{2} + 1$ , $x$ ≥ 0.

The region $R$ is bounded by the curves $y = f (x)$ , $y = g (x)$ and the lines $y = 0$ , $x = 0$ and $y = 9$ as shown on the following diagram.

The shape of a clay vase can be modelled by rotating the region $R$ through 360˚ about the $y$ -axis.

Find the volume of clay used to make the vase.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

volume $= π \int_{0}^{9} {(y^{\frac{1}{2}} + 1)}^{2} d y - π \int_{1}^{9} (y - 1) d y$ (M1)(M1)(M1)(A1)(A1)

Note: Award (M1) for use of formula for rotating about $y$ -axis, (M1) for finding at least one inverse, (M1) for subtracting volumes, (A1)(A1)for each correct expression, including limits.

$= 268.6 \dots - 100.5 \dots (85.5 π - 32 π)$

$= 168 (= 53.5 π)$ A2

[7 marks]

Examiners report

[N/A]

19M.2.AHL.TZ2.H_9b

Question

Consider the polynomial $P (z) \equiv z^{4} - 6 z^{3} - 2 z^{2} + 58 z - 51, z \in C$ .

Sketch the graph of $y = x^{4} - 6 x^{3} - 2 x^{2} + 58 x - 51$ , stating clearly the coordinates of any maximum and minimum points and intersections with axes.

[6]

Hence, or otherwise, state the condition on $k \in R$ such that all roots of the equation $P (z) = k$ are real.

[2]

Markscheme

shape A1

$x$ -axis intercepts at (−3, 0), (1, 0) and $y$ -axis intercept at (0, −51) A1A1

minimum points at (−1.62, −118) and (3.72, 19.7) A1A1

maximum point at (2.40, 26.9) A1

Note: Coordinates may be seen on the graph or elsewhere.

Note: Accept −3, 1 and −51 marked on the axes.

[6 marks]

from graph, 19.7 ≤ $k$ ≤ 26.9 A1A1

Note: Award A1 for correct endpoints and A1 for correct inequalities.

[2 marks]

Examiners report

[N/A]

19M.2.AHL.TZ1.H_10a

Question

The voltage $v$ in a circuit is given by the equation

$v (t) = 3 sin (100 π t)$ , $t ⩾ 0$ where $t$ is measured in seconds.

The current $i$ in this circuit is given by the equation

$i (t) = 2 sin (100 π (t + 0.003))$ .

The power $p$ in this circuit is given by $p (t) = v (t) \times i (t)$ .

The average power $p_{a v}$ in this circuit from $t = 0$ to $t = T$ is given by the equation

$p_{a v} (T) = \frac{1}{T} \int_{0}^{T} p (t) d t$ , where $T > 0$ .

Write down the maximum and minimum value of $v$ .

[2]

Write down two transformations that will transform the graph of $y = v (t)$ onto the graph of $y = i (t)$ .

[2]

Sketch the graph of $y = p (t)$ for 0 ≤ $t$ ≤ 0.02 , showing clearly the coordinates of the first maximum and the first minimum.

[3]

Find the total time in the interval 0 ≤ $t$ ≤ 0.02 for which $p (t)$ ≥ 3.

[3]

Find $p_{a v}$ (0.007).

[2]

With reference to your graph of $y = p (t)$ explain why $p_{a v} (T)$ > 0 for all $T$ > 0.

[2]

Given that $p (t)$ can be written as $p (t) = a sin (b (t - c)) + d$ where $a$ , $b$ , $c$ , $d$ > 0, use your graph to find the values of $a$ , $b$ , $c$ and $d$ .

[6]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

3, −3 A1A1

[2 marks]

stretch parallel to the $y$ -axis (with $x$ -axis invariant), scale factor $\frac{2}{3}$ A1

translation of $(\begin{matrix} - 0.003 \\ 0 \end{matrix})$ (shift to the left by 0.003) A1

Note: Can be done in either order.

[2 marks]

correct shape over correct domain with correct endpoints    A1
first maximum at (0.0035, 4.76)    A1
first minimum at (0.0085, −1.24)    A1

[3 marks]

$p$ ≥ 3 between $t$ = 0.0016762 and 0.0053238 and $t$ = 0.011676 and 0.015324 (M1)(A1)

Note: Award M1A1 for either interval.

= 0.00730 A1

[3 marks]

$p_{a v} = \frac{1}{0.007} \int_{0}^{0.007} 6 sin (100 π t) sin (100 π (t + 0.003)) d t$ (M1)

= 2.87 A1

[2 marks]

in each cycle the area under the $t$ axis is smaller than area above the $t$ axis R1

the curve begins with the positive part of the cycle R1

[2 marks]

$a = \frac{4.76 - (- 1.24)}{2}$ (M1)

$a = 3.00$ A1

$d = \frac{4.76 + (- 1.24)}{2}$

$d = 1.76$ A1

$b = \frac{2 π}{0.01}$

$b = 628 (= 200 π)$ A1

$c = 0.0035 - \frac{0.01}{4}$ (M1)

$c = 0.00100$ A1

[6 marks]

Examiners report

[N/A]

19N.2.AHL.TZ0.H_9c

Question

A body moves in a straight line such that its velocity, $v m s^{- 1}$ , after $t$ seconds is given by $v = 2 sin (\frac{t}{10} + \frac{π}{5}) \csc (\frac{t}{30} + \frac{π}{4})$ for $0 ⩽ t ⩽ 60$ .

The following diagram shows the graph of $v$ against $t$ . Point $A$ is a local maximum and point $B$ is a local minimum.

The body first comes to rest at time $t = t_{1}$ . Find

Determine the coordinates of point $A$ and the coordinates of point $B$ .

[4]

a.i.

Hence, write down the maximum speed of the body.

[1]

a.ii.

the value of $t_{1}$ .

[2]

b.i.

the distance travelled between $t = 0$ and $t = t_{1}$ .

[2]

b.ii.

the acceleration when $t = t_{1}$ .

[2]

b.iii.

Find the distance travelled in the first 30 seconds.

[3]

Markscheme

$A (7.47, 2.28)$ and $B (43.4, - 2.45)$ A1A1A1A1

[4 marks]

a.i.

maximum speed is $2.45 (m s^{- 1})$ A1

[1 mark]

a.ii.

$v = 0 \Rightarrow t_{1} = 25.1 (s)$ (M1)A1

[2 marks]

b.i.

$\int_{0}^{t_{1}} v d t$ (M1)

$= 41.0 (m)$ A1

[2 marks]

b.ii.

$a = \frac{d v}{d t}$ at $t = t_{1} = 25.1$ (M1)

$a = - 0.200 (m s^{- 2})$ A1

Note: Accept $a = - 0.2$ .

[2 marks]

b.iii.

attempt to integrate between 0 and 30 (M1)

Note: An unsupported answer of 38.6 can imply integrating from 0 to 30.

EITHER

$\int_{0}^{30} | v | d t$ (A1)

$41.0 - \int_{t_{1}}^{30} v d t$ (A1)

THEN

$= 43.3 (m)$ A1

[3 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

[N/A]

19N.2.AHL.TZ0.H_11b

Question

The following diagram shows part of the graph of $2 x^{2} = si n^{3} y$ for $0 ⩽ y ⩽ π$ .

The shaded region $R$ is the area bounded by the curve, the $y$ -axis and the lines $y = 0$ and $y = π$ .

Using implicit differentiation, find an expression for $\frac{d y}{d x}$ .

[4]

a.i.

Find the equation of the tangent to the curve at the point $(\frac{1}{4}, \frac{5 π}{6})$ .

[4]

a.ii.

Find the area of $R$ .

[3]

The region $R$ is now rotated about the $y$ -axis, through $2 π$ radians, to form a solid.

By writing $si n^{3} y$ as $(1 - co s^{2} y) sin y$ , show that the volume of the solid formed is $\frac{2 π}{3}$ .

[6]

Markscheme

valid attempt to differentiate implicitly (M1)

$4 x = 3 si n^{2} y cos y \frac{d y}{d x}$ A1A1

$\frac{d y}{d x} = \frac{4 x}{3 si n^{2} y cos y}$ A1

[4 marks]

a.i.

at $(\frac{1}{4}, \frac{5 π}{6}), \frac{d y}{d x} = \frac{4 x}{3 si n^{2} y cos y} = \frac{1}{3 {(\frac{1}{2})}^{2} (- \frac{\sqrt{3}}{2})}$ (M1)

$\Rightarrow \frac{d y}{d x} = - \frac{8}{3 \sqrt{3}} (= - 1.54)$ A1

hence equation of tangent is

$y - \frac{5 π}{6} = - 1.54 (x - \frac{1}{4})$ OR $y = - 1.54 x + 3.00$ (M1)A1

Note: Accept $y = - 1.54 x + 3$ .

[4 marks]

a.ii.

$x = \sqrt{\frac{1}{2} si n^{3} y}$ (M1)

$\int_{0}^{π} \sqrt{\frac{1}{2} si n^{3} y d y}$ (A1)

$= 1.24$ A1

[3 marks]

use of volume $= \int π x^{2} d y$ (M1)

$= \int_{0}^{π} \frac{1}{2} π si n^{3} y d y$ A1

$= \frac{1}{2} π \int_{0}^{π} (sin y - sin y co s^{2} y) d y$

Note: Condone absence of limits up to this point.

reasonable attempt to integrate (M1)

$= \frac{1}{2} π {[- cos y + \frac{1}{3} co s^{3} y]}_{0}^{π}$ A1A1

Note: Award A1 for correct limits (not to be awarded if previous M1 has not been awarded) and A1 for correct integrand.

$= \frac{1}{2} π (1 - \frac{1}{3}) - \frac{1}{2} π (- 1 + \frac{1}{3})$ A1

$= \frac{2 π}{3}$ AG

Note: Do not accept decimal answer equivalent to $\frac{2 π}{3}$ .

[6 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

20N.2.AHL.TZ0.F_5a

Question

Assuming the Maclaurin series for $\cos x$ and $\ln (1 + x)$ , show that the Maclaurin series for $\cos (\ln (1 + x))$ is

$1 - \frac{1}{2} x^{2} + \frac{1}{2} x^{3} - \frac{5}{12} x^{4} + \dots$

[4]

By differentiating the series in part (a), show that the Maclaurin series for $\sin (\ln (1 + x))$ is $x - \frac{1}{2} x^{2} + \frac{1}{6} x^{3} + \dots$ .

[4]

Hence determine the Maclaurin series for $\tan (\ln (1 + x))$ as far as the term in $x^{3}$ .

[5]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

attempts to substitute $\ln (x + 1) = x - \frac{1}{2} x^{2} + \frac{1}{3} x^{3} - \frac{1}{4} x^{4} + \dots$ into

$\cos x = 1 - \frac{1}{2} x^{2} + \frac{1}{24} x^{4} - \dots$ M1

$\cos x = (\ln (1 + x)) = 1 - \frac{1}{2} {(x - \frac{1}{2} x^{2} + \frac{1}{3} x^{3} + \dots)}^{2} + \frac{1}{24} {(x + \dots)}^{4} + \dots$ A1

attempts to expand the $RHS$ up to and including the $x^{4}$ term M1

$= 1 - \frac{1}{2} (x^{2} - x^{3} + \frac{1}{4} x^{4} + \frac{2}{3} x^{4} \dots) + \frac{1}{24} x^{4} + \dots$ A1

$= 1 - \frac{1}{2} x^{2} + \frac{1}{2} x^{3} - \frac{5}{12} x^{4} + \dots$ AG

METHOD 2

attempts to substitute $\ln (x + 1)$ into $\cos x = 1 - \frac{1}{2} x^{2} + \frac{1}{24} x^{4} - \dots$ M1

$\cos (\ln (1 + x)) = 1 - \frac{1}{2} {(\ln (1 + x))}^{2} + \frac{1}{24} {(\ln (1 + x))}^{4} - \dots$

attempts to find the Maclaurin series for ${(\ln (1 + x))}^{2}$ up to and including the $x^{4}$ term M1

${(\ln (1 + x))}^{2} = x^{2} - x^{3} + \frac{11}{12} x^{4} - \dots$ A1

${(\ln (1 + x))}^{2} = x^{4} - \dots$

$= 1 - \frac{1}{2} (x^{2} - x^{3} + \frac{11}{12} x^{4} + \dots) + \frac{1}{24} x^{4} + \dots$ A1

$= 1 - \frac{1}{2} x^{2} + \frac{1}{2} x^{3} - \frac{5}{12} x^{4} + \dots$ AG

[4 marks]

$- \sin (\ln (1 + x)) \times \frac{1}{1 + x} = - x + \frac{3}{2} x^{2} - \frac{5}{3} x^{3} + \dots$ A1A1

$\sin (\ln (1 + x)) = - (1 + x) (- x + \frac{3}{2} x^{2} - \frac{5}{3} x^{3} + \dots)$

attempts to expand the $RHS$ up to and including the $x^{3}$ term M1

$= x - \frac{3}{2} x^{2} + \frac{5}{3} x^{3} + x^{2} - \frac{3}{2} x^{3} + \dots$ A1

$= x - \frac{1}{2} x^{2} + \frac{1}{6} x^{3} + \dots$ AG

[4 marks]

METHOD 1

let $\tan (\ln (1 + x)) = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + \dots$

uses $\sin (\ln (1 + x)) = \cos (\ln (1 + x)) \times \tan (\ln (1 + x))$ to form M1

$x - \frac{1}{2} x^{2} + \frac{1}{6} x^{3} + \dots = (1 - \frac{1}{2} x^{2} + \frac{1}{2} x^{3} + \dots) (a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + \dots)$ A1

$= a_{0} + a_{1} x + (a_{2} - \frac{1}{2} a_{0}) x^{2} + (a_{3} - \frac{1}{2} a_{1} + \frac{1}{2} a_{0}) x^{3} + \dots$ (A1)

attempts to equate coefficients,

$a_{0} = 0, a_{1} = 1, a_{2} - \frac{1}{2} a_{0} = - \frac{1}{2}, a_{3} - \frac{1}{2} a_{1} + \frac{1}{2} a_{0} = \frac{1}{6}$ M1

$a_{0} = 0, a_{1} = 1, a_{2} = - \frac{1}{2}, a_{3} = \frac{2}{3}$ A1

so $\tan (\ln (1 + x)) = x - \frac{1}{2} x^{2} + \frac{2}{3} x^{3} + \dots$

METHOD 2

uses $\tan (\ln (1 + x)) = \frac{\sin (\ln (1 + x))}{\cos (\ln (1 + x))}$ to form M1

$= (x - \frac{1}{2} x^{2} + \frac{1}{6} x^{3} + \dots) {(1 - \frac{1}{2} x^{2} + \frac{1}{2} x^{3} + \dots)}^{- 1}$ A1

$= {(1 - \frac{1}{2} x^{2} + \frac{1}{2} x^{3} + \dots)}^{- 1} = 1 + \frac{1}{2} x^{2} - \frac{1}{2} x^{3} + \dots$ (A1)

attempts to expand the $RHS$ up to and including the $x^{3}$ term M1

$= (x - \frac{1}{2} x^{2} + \frac{1}{6} x^{3} + \dots) (1 + \frac{1}{2} x^{2} - \frac{1}{2} x^{3} + \dots)$

$= x + \frac{1}{2} x^{3} - \frac{1}{2} x^{2} + \frac{1}{6} x^{3} + \dots$

$= x - \frac{1}{2} x^{2} + \frac{2}{3} x^{3} + \dots$ A1

Note: Accept use of long division.

[5 marks]

Examiners report

[N/A]

20N.2.AHL.TZ0.F_9a

Question

Consider the differential equation $\frac{d y}{d x} = \frac{y - x}{y + x}$ , where $x, y > 0$ .

It is given that $y = 2$ when $x = 1$ .

Solve the differential equation, giving your answer in the form $f (x, y) = 0$ .

[9]

The graph of $y$ against $x$ has a local maximum between $x = 2$ and $x = 3$ . Determine the coordinates of this local maximum.

[4]

Show that there are no points of inflexion on the graph of $y$ against $x$ .

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

puts $y = v x$ so that $\frac{d y}{d x} = v + x \frac{d v}{d x}$ M1

$v + x \frac{d v}{d x} = \frac{v x - x}{v x + x} (= \frac{v - 1}{v + 1})$ A1

attempts to express $x \frac{d v}{d x}$ as a single rational fraction in $v$

$x \frac{d v}{d x} = - \frac{v^{2} + 1}{v + 1}$ M1

attempts to separate variables M1

$\int \frac{v + 1}{v^{2} + 1} d v = - \int \frac{1}{x} d x$

$\frac{1}{2} \ln (v^{2} + 1) + arctan v = - \ln x (+ C)$ A1A1

substitutes $y = 2, x = 1$ and attempts to find the value of $C$ M1

$C = \frac{1}{2} \ln 5 + arctan 2$ A1

the solution is

$\frac{1}{2} \ln (\frac{y^{2}}{x^{2}} + 1) + arctan (\frac{y}{x}) + \ln x - \frac{1}{2} \ln 5 - arctan 2 = 0$ A1

[9 marks]

at a maximum, $\frac{d y}{d x} = 0$ M1

attempts to substitute $y = x$ into their solution M1

$\frac{1}{2} \ln 2 + arctan 1 + \ln x = \frac{1}{2} \ln 5 + arctan 2$

attempts to solve for $x, y$ (M1)

$(2.18, 2.18) (\frac{\sqrt{10}}{2} e^{arctan 2 - \frac{π}{4}}, \frac{\sqrt{10}}{2} e^{arctan 2 - \frac{π}{4}})$ A1

Note: Accept all answers that round to the correct $2 sf$ answer.
Accept $x = 2.18, y = 2.18$ .

[4 marks]

METHOD 1

attempts (quotient rule) implicit differentiation M1

$\frac{d^{2} y}{d x^{2}} = \frac{(\frac{d y}{d x} - 1) (y + x) - (y - x) (\frac{d y}{d x} + 1)}{{(y + x)}^{2}}$

correctly substitutes $\frac{d y}{d x} = \frac{y - x}{y + x}$ into $\frac{d^{2} y}{d x^{2}}$

$= \frac{(\frac{y - x}{y + x} - 1) (y + x) - (y - x) (\frac{y - x}{y + x} + 1)}{{(y + x)}^{2}}$ A1

$= - \frac{2 (x^{2} + y^{2})}{{(x + y)}^{3}}$ A1

this expression can never be zero therefore no points of inflexion R1

METHOD 2

attempts implicit differentiation on $(y + x) \frac{d y}{d x} = y - x$ M1

$(\frac{d y}{d x} + 1) \frac{d y}{d x} + (y + x) \frac{d^{2} y}{d x^{2}} = \frac{d y}{d x} - 1$ A1

$(y + x) \frac{d^{2} y}{d x^{2}} = \frac{d y}{d x} - 1 - {(\frac{d y}{d x})}^{2} - \frac{d y}{d x}$

$= - 1 - {(\frac{d y}{d x})}^{2}$ A1

$- 1 - {(\frac{d y}{d x})}^{2} < 0$ and $x + y > 0, \frac{d^{2} y}{d x^{2}} \neq 0$ therefore no points of inflexion R1

Note: Accept putting $\frac{d^{2} y}{d x^{2}} = 0$ and obtaining contradiction.

[4 marks]

Examiners report

[N/A]

20N.2.AHL.TZ0.H_3a

Question

The following diagram shows part of the graph of $y = p + q \sin (r x)$ . The graph has a local maximum point at $(- \frac{9 π}{4}, 5)$ and a local minimum point at $(- \frac{3 π}{4}, - 1)$ .

Determine the values of $p$ , $q$ and $r$ .

[4]

Hence find the area of the shaded region.

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

the principal axis is $\frac{5 + (- 1)}{2} (= 2)$

so $p = 2$ A1

the amplitude is $\frac{5 - (- 1)}{2} (= 3)$

so $q = 3$ A1

EITHER

one period is $2 (- \frac{3 π}{4} - (- \frac{9 π}{4}))$ (M1)

$= 3 π$

$\Rightarrow \frac{2 π}{r} = 3 π$

Substituting a point eg $- 1 = 2 + \sin (- \frac{3 π}{4} r)$

$\sin (- \frac{3 π}{4} r) = - 1 \Rightarrow - \frac{3 π}{4} r = \dots - \frac{5 π}{2}, - \frac{π}{2}, \frac{3 π}{2}, \dots$

Choice of correct solution $- \frac{3 π}{4} r = - \frac{π}{2}$ (M1)

THEN

$\Rightarrow r = \frac{2}{3}$ A1

$(\Rightarrow y = 2 + 3 \sin (\frac{2 x}{3}))$

Note: $q$ and $r$ can be both given as negatives for full marks

[4 marks]

roots are $x = - 1.09459 \dots, x = - 3.617797 \dots$ (A1)

$\int_{- 3.617797 \dots}^{- 1.09459 \dots} (2 + 3 \sin (\frac{2 x}{3})) d x$ (M1)

$= - 1.66 (= - 1.66179 \dots)$ (A1)

so area $= 1.66 ({units}^{2})$ A1

[4 marks]

Examiners report

[N/A]

20N.2.AHL.TZ0.H_8a

Question

A small bead is free to move along a smooth wire in the shape of the curve $y = \frac{10}{3 - 2 e^{- 0.5 x}} (x \geq 0)$ .

Find an expression for $\frac{d y}{d x}$ .

[3]

At the point on the curve where $x = 4$ , it is given that $\frac{d y}{d t} = - 0.1 {m s}^{- 1}$

Find the value of $\frac{d x}{d t}$ at this exact same instant.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid attempt to use chain rule or quotient rule (M1)

$\frac{d y}{d x} = \frac{- 10 e^{- 0.5 x}}{{(3 - 2 e^{- 0.5 x})}^{2}}$ OR $\frac{d y}{d x} = - 10 e^{- 0.5 x} {(3 - 2 e^{- 0.5 x})}^{- 2}$ A1A1

[3 marks]

Note: Award A1 for numerator and A1 for denominator, or A1 for each part if the second alternative given.

valid attempt to use chain rule $(eg \frac{d y}{d t} = \frac{d y}{d x} \times \frac{d x}{d t})$ (M1)

$\frac{d x}{d t} = - 0.1 \div \frac{- 10 e^{- 2}}{{(3 - 2 e^{- 2})}^{2}} (= - 0.1 \div - 0.181676 \dots)$ or equivalent (A1)

$= 0.550428 \dots$

$\frac{d x}{d t} = 0.550 ({ms}^{- 1})$ A1

[3 marks]

Examiners report

[N/A]

20N.2.AHL.TZ0.H_11a

Question

A particle $P$ moves in a straight line such that after time $t$ seconds, its velocity, $v$ in ${m s}^{- 1}$ , is given by $v = e^{- 3 t} \sin 6 t$ , where $0 < t < \frac{π}{2}$ .

At time $t$ , $P$ has displacement $s (t)$ ; at time $t = 0$ , $s (0) = 0$ .

At successive times when the acceleration of $P$ is $0 {m s}^{- 2}$ , the velocities of $P$ form a geometric sequence. The acceleration of $P$ is zero at times $t_{1}, t_{2}, t_{3}$ where $t_{1} < t_{2} < t_{3}$ and the respective velocities are $v_{1}, v_{2}, v_{3}$ .

Find the times when $P$ comes to instantaneous rest.

[2]

Find an expression for $s$ in terms of $t$ .

[7]

Find the maximum displacement of $P$ , in metres, from its initial position.

[2]

Find the total distance travelled by $P$ in the first $1.5$ seconds of its motion.

[2]

Show that, at these times, $\tan 6 t = 2$ .

[2]

e.i.

Hence show that $\frac{v_{2}}{v_{1}} = \frac{v_{3}}{v_{2}} = - e^{- \frac{π}{2}}$ .

[5]

e.ii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\frac{π}{6} (= 0.524)$ A1

$\frac{π}{3} (= 1.05)$ A1

[2 marks]

attempt to use integration by parts M1

$s = \int e^{- 3 t} \sin 6 t d t$

EITHER

$= - \frac{e^{- 3 t} \sin 6 t}{3} - \int - 2 e^{- 3 t} cos 6 t d t$ A1

$= - \frac{e^{- 3 t} \sin 6 t}{3} - (\frac{2 e^{- 3 t} cos 6 t}{3} - \int - 4 e^{- 3 t} sin 6 t d t)$ A1

$= - \frac{e^{- 3 t} \sin 6 t}{3} - (\frac{2 e^{- 3 t} cos 6 t}{3} + 4 s)$

$5 s = \frac{-3 e^{- 3 t} \sin 6 t - 6 e^{- 3 t} cos 6 t}{9}$ M1

$= - \frac{e^{- 3 t} cos 6 t}{6} - \int \frac{1}{2} e^{- 3 t} cos 6 t d t$ A1

$= - \frac{e^{- 3 t} cos 6 t}{6} - (\frac{e^{- 3 t} sin 6 t}{12} + \int \frac{1}{4} e^{- 3 t} sin 6 t d t)$ A1

$= - \frac{e^{- 3 t} cos 6 t}{6} - (\frac{e^{- 3 t} sin 6 t}{12} + \frac{1}{4} s)$

$\frac{5}{4} s = \frac{- 2 e^{- 3 t} cos 6 t - e^{- 3 t} sin 6 t}{12}$ M1

THEN

$s = - \frac{e^{- 3 t} (sin 6 t + 2 cos 6 t)}{15} (+ c)$ A1

at $t = 0, s = 0 \Rightarrow 0 = - \frac{2}{15} + c$ M1

$c = \frac{2}{15}$ A1

$s = \frac{2}{15} - \frac{e^{- 3 t} (sin 6 t + 2 cos 6 t)}{15}$

[7 marks]

EITHER

substituting $t = \frac{π}{6}$ into their equation for $s$ (M1)

$(s = \frac{2}{15} - \frac{e^{- \frac{π}{2}} (sin π + 2 cos π)}{15})$

using GDC to find maximum value (M1)

evaluating $\int_{0}^{\frac{π}{6}} v d t$ (M1)

THEN

$= 0.161 (= \frac{2}{15} (1 + e^{- \frac{π}{2}}))$ A1

[2 marks]

METHOD 1

EITHER

distance required $= \int_{0}^{1.5} |e^{- 3 t} \sin 6 t| d t$ (M1)

distance required $= \int_{0}^{\frac{π}{6}} e^{- 3 t} \sin 6 t d t + |\int_{\frac{π}{6}}^{\frac{π}{3}} e^{- 3 t} \sin 6 t d t| + \int_{\frac{π}{3}}^{1.5} e^{- 3 t} \sin 6 t d t$ (M1)

$(= 0.16105 \dots + 0.033479 \dots + 0.006806 \dots)$

THEN

$= 0.201 (m)$ A1

METHOD 2

using successive minimum and maximum values on the displacement graph (M1)

$0.16105 \dots + (0.16105 \dots - 0.12757 \dots) + (0.13453 \dots - 0.12757 \dots)$

$= 0.201 (m)$ A1

[2 marks]

valid attempt to find $\frac{d v}{d t}$ using product rule and set $\frac{d v}{d t} = 0$ M1

$\frac{d v}{d t} = e^{- 3 t} 6 \cos 6 t - 3 e^{- 3 t} \sin 6 t$ A1

$\frac{d v}{d t} = 0 \Rightarrow \tan 6 t = 2$ AG

[2 marks]

e.i.

attempt to evaluate $t_{1}, t_{2}, t_{3}$ in exact form M1

$6 t_{1} = arctan 2 (\Rightarrow t_{1} = \frac{1}{6} arctan 2)$

$6 t_{2} = π + arctan 2 (\Rightarrow t_{2} = \frac{π}{6} + \frac{1}{6} arctan 2)$

$6 t_{3} = 2 π + arctan 2 (\Rightarrow t_{3} = \frac{π}{3} + \frac{1}{6} arctan 2)$ A1

Note: The A1 is for any two consecutive correct, or showing that $6 t_{2} = π + 6 t_{1}$ or $6 t_{3} = π + 6 t_{2}$ .

showing that $\sin 6 t_{n + 1} = - \sin 6 t_{n}$

eg $\tan 6 t = 2 \Rightarrow \sin 6 t = \pm \frac{2}{\sqrt{5}}$ M1A1

showing that $\frac{e^{- 3 t_{n +1}}}{e^{- 3 t_{n}}} = e^{- \frac{π}{2}}$ M1

eg $e^{- 3 (\frac{π}{6} + k)} \div e^{- 3 k} = e^{- \frac{π}{2}}$

Note: Award the A1 for any two consecutive terms.

$\frac{v_{3}}{v_{2}} = \frac{v_{2}}{v_{1}} = - e^{- \frac{π}{2}}$ AG

[5 marks]

e.ii.

Examiners report

[N/A]

e.i.

[N/A]

e.ii.

21M.2.AHL.TZ1.7a

Question

A continuous random variable $X$ has the probability density function $f$ given by

$f (x) = \{\begin{cases} \frac{x}{\sqrt{{(x^{2} + k)}^{3}}} 0 \leq x \leq 4 \\ 0 otherwise \end{cases}$

where $k \in ℝ^{+}$ .

Show that $\sqrt{16 + k} - \sqrt{k} = \sqrt{k} \sqrt{16 + k}$ .

[5]

Find the value of $k$ .

[2]

Markscheme

recognition of the need to integrate $\frac{x}{\sqrt{{(x^{2} + k)}^{3}}}$ (M1)

$\int \frac{x}{\sqrt{{(x^{2} + k)}^{3}}} d x (= 1)$

EITHER

$u = x^{2} + k \Rightarrow \frac{d u}{d x} = 2 x$ (or equivalent) (A1)

$\int \frac{x}{\sqrt{{(x^{2} + k)}^{3}}} d x = \frac{1}{2} \int u^{- \frac{3}{2}} d u$

$= - u^{- \frac{1}{2}} (+ c) (= - {(x^{2} + k)}^{- \frac{1}{2}} (+ c))$ A1

$\int \frac{x}{\sqrt{{(x^{2} + k)}^{3}}} d x = \frac{1}{2} \int \frac{2 x}{\sqrt{{(x^{2} + k)}^{3}}} d x$ (A1)

$= - {(x^{2} + k)}^{- \frac{1}{2}} (+ c)$ A1

THEN

attempt to use correct limits for their integrand and set equal to $1$ M1

${[- u^{- \frac{1}{2}}]}_{k}^{16 + k} = 1$ OR ${[- {(x^{2} + k)}^{- \frac{1}{2}}]}_{0}^{4} = 1$

$- {(16 + k)}^{- \frac{1}{2}} + k^{- \frac{1}{2}} = 1 (\Rightarrow \frac{1}{\sqrt{k}} - \frac{1}{\sqrt{16 + k}} = 1)$ A1

$\sqrt{16 + k} - \sqrt{k} = \sqrt{k} \sqrt{16 + k}$ AG

[5 marks]

attempt to solve $\sqrt{16 + k} - \sqrt{k} = \sqrt{k} \sqrt{16 + k}$ (M1)

$k = 0.645038 \dots$

$= 0.645$ A1

[2 marks]

Examiners report

[N/A]

21M.2.AHL.TZ2.9a

Question

Write down the first three terms of the binomial expansion of $(1 + t)^{- 1}$ in ascending powers of $t$ .

[1]

By using the Maclaurin series for $\cos x$ and the result from part (a), show that the Maclaurin series for $sec x$ up to and including the term in $x^{4}$ is $1 + \frac{x^{2}}{2} + \frac{5 x^{4}}{24}$ .

[4]

By using the Maclaurin series for $arctan x$ and the result from part (b), find $\lim_{x \to 0} (\frac{x arctan 2 x}{sec x - 1})$ .

[3]

Markscheme

$1 - t + t^{2}$ A1

Note: Accept $1, - t$ and $t^{2}$ .

[1 mark]

$sec x = \frac{1}{1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} (- \dots)} (= {(1 - \frac{x^{2}}{2!} + (\frac{x^{4}}{4!} (- \dots)))}^{- 1})$ (M1)

$t = \cos x - 1$ or $sec x = 1 - (\cos x - 1) + {(\cos x - 1)}^{2}$ (M1)

$= 1 - (- \frac{x^{2}}{2!} + \frac{x^{4}}{4!} (- \dots)) + {(- \frac{x^{2}}{2!} + \frac{x^{4}}{4!} (- \dots))}^{2}$ A1

$= 1 + \frac{x^{2}}{2} - \frac{x^{4}}{24} + \frac{x^{4}}{4}$ A1

so the Maclaurin series for $sec x$ up to and including the term in $x^{4}$ is $1 + \frac{x^{2}}{2} + \frac{5 x^{4}}{24}$ AG

Note: Condone the absence of ‘…’

[4 marks]

$arctan 2 x = 2 x - \frac{{(2 x)}^{3}}{3} + \dots$

$\lim_{x \to 0} (\frac{x arctan 2 x}{sec x - 1}) = \lim_{x \to 0} (\frac{x (2 x - \frac{{(2 x)}^{3}}{3} + \dots)}{(1 + \frac{x^{2}}{2} + \frac{5 x^{4}}{24}) - 1})$ M1

$= \lim_{x \to 0} (\frac{2 x^{2} - \frac{8 x^{4}}{3} + \dots}{\frac{x^{2}}{2} + \frac{5 x^{4}}{24}})$ A1

$= \lim_{x \to 0} (\frac{2 x^{2} (1 - \frac{4 x^{2}}{3})}{\frac{x^{2}}{2} (1 + \frac{5 x^{2}}{12})})$

$= 4$ A1

Note: Condone missing ‘lim’ and errors in higher derivatives.
Do not award M1 unless $x$ is replaced by $2 x$ in $arctan$ .

[3 marks]

Examiners report

[N/A]

21M.2.AHL.TZ1.9b

Question

Two boats $A$ and $B$ travel due north.

Initially, boat $B$ is positioned $50$ metres due east of boat $A$ .

The distances travelled by boat $A$ and boat $B$ , after $t$ seconds, are $x$ metres and $y$ metres respectively. The angle $θ$ is the radian measure of the bearing of boat $B$ from boat $A$ . This information is shown on the following diagram.

Show that $y = x + 50 cot θ$ .

[1]

At time $T$ , the following conditions are true.

Boat $B$ has travelled $10$ metres further than boat $A$ .
Boat $B$ is travelling at double the speed of boat $A$ .
The rate of change of the angle $θ$ is $- 0.1$ radians per second.

Find the speed of boat $A$ at time $T$ .

[6]

Markscheme

$\tan θ = \frac{50}{y - x}$ OR $cot θ = \frac{y - x}{50}$ A1

$y = x + 50 cot θ$ AG

Note: $y - x$ may be identified as a length on a diagram, and not written explicitly.

[1 mark]

attempt to differentiate with respect to $t$ (M1)

$\frac{d y}{d t} = \frac{d x}{d t} - 50 {(cosec θ)}^{2} \frac{d θ}{d t}$ A1

attempt to set speed of $B$ equal to double the speed of $A$ (M1)

$2 \frac{d x}{d t} = \frac{d x}{d t} - 50 {(cosec θ)}^{2} \frac{d θ}{d t}$

$\frac{d x}{d t} = - 50 {(cosec θ)}^{2} \frac{d θ}{d t}$ A1

$θ = arctan 5 (= 1.373 \dots = 78.69 \dots °)$ OR ${cosec}^{2} θ = 1 + {cot}^{2} θ = 1 + {(\frac{1}{5})}^{2} = \frac{26}{25}$ (A1)

Note: This A1 can be awarded independently of previous marks.

$\frac{d x}{d t} = - 50 (\frac{26}{25}) \times - 0.1$

So the speed of boat $A$ is $5.2 ({ms}^{- 1})$ A1

Note: Accept $5.20$ from the use of inexact values.

[6 marks]

Examiners report

[N/A]

21M.2.AHL.TZ2.11a

Question

A function $f$ is defined by $f (x) = \frac{k e^{\frac{x}{2}}}{1 + e^{x}}$ , where $x \in ℝ, x \geq 0$ and $k \in ℝ^{+}$ .

The region enclosed by the graph of $y = f (x)$ , the $x$ -axis, the $y$ -axis and the line $x = \ln 16$ is rotated $360 °$ about the $x$ -axis to form a solid of revolution.

Pedro wants to make a small bowl with a volume of $300 {cm}^{3}$ based on the result from part (a). Pedro’s design is shown in the following diagrams.

The vertical height of the bowl, $BO$ , is measured along the $x$ -axis. The radius of the bowl’s top is $OA$ and the radius of the bowl’s base is $BC$ . All lengths are measured in $cm$ .

For design purposes, Pedro investigates how the cross-sectional radius of the bowl changes.

Show that the volume of the solid formed is $\frac{15 k^{2} π}{34}$ cubic units.

[6]

Find the value of $k$ that satisfies the requirements of Pedro’s design.

[2]

Find $OA$ .

[2]

c.i.

Find $BC$ .

[2]

c.ii.

By sketching the graph of a suitable derivative of $f$ , find where the cross-sectional radius of the bowl is decreasing most rapidly.

[4]

d.i.

State the cross-sectional radius of the bowl at this point.

[2]

d.ii.

Markscheme

attempt to use $V = π \int_{a}^{b} {(f (x))}^{2} d x$ (M1)

$V = π \int_{0}^{\ln 16} {(\frac{k e^{\frac{x}{2}}}{1 + e^{x}})}^{2} d x (V = k^{2} π \int_{0}^{\ln 16} \frac{e^{x}}{{(1 + e^{x})}^{2}} d x)$

EITHER

applying integration by recognition (M1)

$= k^{2} π {[- \frac{1}{1 + e^{x}}]}_{0}^{\ln 16}$ A3

$u = 1 + e^{x} \Rightarrow d u = e^{x} d x$ (A1)

attempt to express the integral in terms of $u$ (M1)

when $x = 0, u = 2$ and when $x = \ln 16, u = 17$

$V = k^{2} π \int_{2}^{17} \frac{1}{u^{2}} d u$ (A1)

$= k^{2} π {[- \frac{1}{u}]}_{2}^{17}$ A1

$u = e^{x} \Rightarrow d u = e^{x} d x$ (A1)

attempt to express the integral in terms of $u$ (M1)

when $x = 0, u = 1$ and when $x = \ln 16, u = 16$

$V = k^{2} π \int_{1}^{16} \frac{1}{{(1 + u)}^{2}} d u$ (A1)

$= k^{2} π {[- \frac{1}{1 + u}]}_{1}^{16}$ A1

Note: Accept equivalent working with indefinite integrals and original limits for $x$ .

THEN

$= k^{2} π (\frac{1}{2} - \frac{1}{17})$ A1

so the volume of the solid formed is $\frac{15 k^{2} π}{34}$ cubic units AG

Note: Award (M1)(A0)(M0)(A0)(A0)(A1) when $\frac{15}{34}$ is obtained from GDC

[6 marks]

a valid algebraic or graphical attempt to find $k$ (M1)

$k^{2} = \frac{300 \times 34}{15 π}$

$k = 14.7 (= 2 \sqrt{\frac{170}{π}} = \sqrt{\frac{680}{π}})$ (as $k \in ℝ^{+}$ ) A1

Note: Candidates may use their GDC numerical solve feature.

[2 marks]

attempting to find $OA = f (0) = \frac{k}{2}$

with $k = 14.712 \dots (= 2 \sqrt{\frac{170}{π}} = \sqrt{\frac{680}{π}})$ (M1)

$OA = 7.36 (= \sqrt{\frac{170}{π}})$ A1

[2 marks]

c.i.

attempting to find $BC = f (\ln 16) = \frac{4 k}{17}$

with $k = 14.712 \dots (= 2 \sqrt{\frac{170}{π}} = \sqrt{\frac{680}{π}})$ (M1)

$BC = 3.46 (= \frac{8}{17} \sqrt{\frac{170}{π}} = \frac{8 \sqrt{10}}{\sqrt{17 π}})$ A1

[2 marks]

c.ii.

EITHER

recognising to graph $y = f' (x)$ (M1)

Note: Award M1 for attempting to use quotient rule or product rule differentiation. $f' (x) = \frac{k e^{\frac{x}{2}} (1 - e^{x})}{2 {(1 + e^{x})}^{2}}$

for $x > 0$ graph decreasing to the local minimum A1

before increasing towards the $x$ -axis A1

recognising to graph $y = f'' (x)$ (M1)

Note: Award M1 for attempting to use quotient rule or product rule differentiation. $f'' (x) = \frac{k e^{\frac{x}{2}} (e^{2 x} - 6 e^{x} +1)}{4 {(1 + e^{x})}^{3}}$

for $x > 0$ , graph increasing towards and beyond the $x$ -intercept A1

recognising $f'' (x) = 0$ for maximum rate (A1)

THEN

$x = 1.76 (= \ln (2 \sqrt{2} + 3))$ A1

Note: Only award A marks if either graph is seen.

[4 marks]

d.i.

attempting to find $f (1.76 \dots)$ (M1)

the cross-sectional radius at this point is $5.20 (\sqrt{\frac{85}{π}}) (cm)$ A1

[2 marks]

d.ii.

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

d.i.

[N/A]

d.ii.

21M.2.AHL.TZ1.11b

Question

The function $f$ is defined by $f (x) = \frac{3 x + 2}{4 x^{2} - 1}$ , for $x \in ℝ$ , $x \neq p$ , $x \neq q$ .

The graph of $y = f (x)$ has exactly one point of inflexion.

The function $g$ is defined by $g (x) = \frac{4 x^{2} - 1}{3 x + 2}$ , for $x \in ℝ, x \neq - \frac{2}{3}$ .

Find the value of $p$ and the value of $q$ .

[2]

Find an expression for $f' (x)$ .

[3]

Find the $x$ -coordinate of the point of inflexion.

[2]

Sketch the graph of $y = f (x)$ for $- 3 \leq x \leq 3$ , showing the values of any axes intercepts, the coordinates of any local maxima and local minima, and giving the equations of any asymptotes.

[5]

Find the equations of all the asymptotes on the graph of $y = g (x)$ .

[4]

By considering the graph of $y = g (x) - f (x)$ , or otherwise, solve $f (x) < g (x)$ for $x \in ℝ$ .

[4]

Markscheme

attempt to solve $4 x^{2} - 1 = 0$ e.g. by factorising $4 x^{2} - 1$ (M1)

$p = \frac{1}{2}, q = - \frac{1}{2}$ or vice versa A1

[2 marks]

attempt to use quotient rule or product rule (M1)

EITHER

$f' (x) = \frac{3 (4 x^{2} - 1) - 8 x (3 x + 2)}{{(4 x^{2} - 1)}^{2}} (= \frac{- 12 x^{2} - 16 x - 3}{{(4 x^{2} - 1)}^{2}})$ A1A1

Note: Award A1 for each term in the numerator with correct signs, provided correct denominator is seen.

$f' (x) = - 8 x (3 x + 2) {(4 x^{2} - 1)}^{- 2} + 3 {(4 x^{2} - 1)}^{- 1}$ A1A1

Note: Award A1 for each term.

[3 marks]

attempt to find the local min point on $y = f' (x)$ OR solve $f'' (x) = 0$ (M1)

$x = - 1.60$ A1

[2 marks]

A1A1A1A1A1

Note: Award A1 for both vertical asymptotes with their equations, award A1 for horizontal asymptote with equation, award A1 for each correct branch including asymptotic behaviour, coordinates of minimum and maximum points (may be seen next to the graph) and values of axes intercepts.
If vertical asymptotes are absent (or not vertical) and the branches overlap as a consequence, award maximum A0A1A0A1A1.

[5 marks]

$x = - \frac{2}{3} (= - 0.667)$ A1

(oblique asymptote has) gradient $\frac{4}{3} (= 1.33)$ (A1)

appropriate method to find complete equation of oblique asymptote M1

$3 x + 2 \overset{\frac{4}{3} x - \frac{8}{9}}{4 x^{2} + 0 x - 1}$

$\frac{4 x^{2} + \frac{8}{3} x}{- \frac{8}{3} x - 1}$

$- \frac{8}{3} x - \frac{16}{\frac{9}{\frac{7}{9}}}$

$y = \frac{4}{3} x - \frac{8}{9} (= 1.33 x - 0.889)$ A1

Note: Do not award the final A1 if the answer is not given as an equation.

[4 marks]

attempting to find at least one critical value $(x = - 0.568729 \dots, x = 1.31872 \dots)$ (M1)

$- \frac{2}{3} < x < - 0.569$ OR $- 0.5 < x < 0.5$ OR $x > 1.32$ A1A1A1

Note: Only penalize once for use of $\leq$ rather than $<$ .

[4 marks]

Examiners report

[N/A]

21M.2.AHL.TZ1.12a

Question

The function $f$ has a derivative given by $f' (x) = \frac{1}{x (k - x)}, x \in ℝ, x \neq o, x \neq k$ where $k$ is a positive constant.

Consider $P$ , the population of a colony of ants, which has an initial value of $1200$ .

The rate of change of the population can be modelled by the differential equation $\frac{d P}{d t} = \frac{P (k - P)}{5 k}$ , where $t$ is the time measured in days, $t \geq 0$ , and $k$ is the upper bound for the population.

At $t = 10$ the population of the colony has doubled in size from its initial value.

The expression for $f' (x)$ can be written in the form $\frac{a}{x} + \frac{b}{k - x}$ , where $a, b \in ℝ$ . Find $a$ and $b$ in terms of $k$ .

[3]

Hence, find an expression for $f (x)$ .

[3]

By solving the differential equation, show that $P = \frac{1200 k}{(k - 1200) e^{- \frac{t}{5}} + 1200}$ .

[8]

Find the value of $k$ , giving your answer correct to four significant figures.

[3]

Find the value of $t$ when the rate of change of the population is at its maximum.

[3]

Markscheme

$\frac{1}{x (k - x)} \equiv \frac{a}{x} + \frac{b}{k - x}$

$a (k - x) + b x = 1$ (A1)

attempt to compare coefficients OR substitute $x = k$ and $x = 0$ and solve (M1)

$a = \frac{1}{k}$ and $b = \frac{1}{k}$ A1

$f' (x) = \frac{1}{k x} + \frac{1}{k (k - x)}$

[3 marks]

attempt to integrate their $\frac{a}{x} + \frac{b}{k - x}$ (M1)

$f (x) \frac{1}{k} \int (\frac{1}{x} + \frac{1}{k - x}) d x$

$= \frac{1}{k} (\ln |x| - \ln |k - x|) (+ c) (= \frac{1}{k} \ln |\frac{x}{k - x}| (+ c))$ A1A1

Note: Award A1 for each correct term. Award A1A0 for a correct answer without modulus signs. Condone the absence of $+ c$ .

[3 marks]

attempt to separate variables and integrate both sides M1

$5 k \int \frac{1}{P (k - P)} d P = \int 1 d t$

$5 (\ln P - \ln (k - P)) = t + c$ A1

Note: There are variations on this which should be accepted, such as $\frac{1}{k} (\ln P - \ln (k - P)) = \frac{1}{5 k} t + c$ . Subsequent marks for these variations should be awarded as appropriate.

EITHER

attempt to substitute $t = 0, P = 1200$ into an equation involving $c$ M1

$c = 5 (\ln 1200 - \ln (k - 1200)) (= 5 \ln (\frac{1200}{k - 1200}))$ A1

$5 (\ln P - \ln (k - P)) = t + 5 (\ln 1200 - \ln (k - 1200))$ A1

$\ln (\frac{P (k - 1200)}{1200 (k - P)}) = \frac{t}{5}$

$\frac{P (k - 1200)}{1200 (k - P)} = e^{\frac{t}{5}}$ A1

$\ln (\frac{P}{k - P}) = \frac{t + c}{5}$

$\frac{P}{k - P} = A e^{\frac{t}{5}}$ A1

attempt to substitute $t = 0, P = 1200$ M1

$\frac{1200}{k - 1200} = A$ A1

$\frac{P}{k - P} = \frac{1200 e^{\frac{t}{5}}}{k - 1200}$ A1

THEN

attempt to rearrange and isolate $P$ M1

$P k - 1200 P = 1200 k e^{\frac{t}{5}} - 1200 P e^{\frac{t}{5}}$ OR $P k e^{- \frac{t}{5}} - 1200 P e^{- \frac{t}{5}} = 1200 k - 1200 P$ OR $\frac{k}{P} - 1 = \frac{k - 1200}{1200 e^{\frac{t}{5}}}$

$P (k - 1200 + 1200 e^{\frac{t}{5}}) = 1200 k e^{\frac{t}{5}}$ OR $P (k e^{- \frac{t}{5}} - 1200 e^{- \frac{t}{5}} + 1200) = 1200 k$ A1

$P = \frac{1200 k}{(k - 1200) e^{- \frac{t}{5}} + 1200}$ AG

[8 marks]

attempt to substitute $t = 10, P = 2400$ (M1)

$2400 = \frac{1200 k}{(k - 1200) e^{- 2} + 1200}$ (A1)

$k = 2845.34 \dots$

$k = 2845$ A1

Note: Award (M1)(A1)A0 for any other value of $k$ which rounds to $2850$

[3 marks]

attempt to find the maximum of the first derivative graph OR zero of the second derivative graph OR that $P = \frac{k}{2} (= 1422.67 \dots)$ (M1)

$t = 1.57814 \dots$

$= 1.58$ (days) A2

Note: Accept any value which rounds to $1.6$ .

[3 marks]

Examiners report

[N/A]

21M.2.AHL.TZ2.12b

Question

A function $f$ is defined by $f (x) = arcsin (\frac{x^{2} - 1}{x^{2} + 1}), x \in ℝ$ .

A function $g$ is defined by $g (x) = arcsin (\frac{x^{2} - 1}{x^{2} + 1}), x \in ℝ, x \geq 0$ .

Show that $f$ is an even function.

[1]

By considering limits, show that the graph of $y = f (x)$ has a horizontal asymptote and state its equation.

[2]

Show that $f' (x) = \frac{2 x}{\sqrt{x^{2}} (x^{2} + 1)}$ for $x \in ℝ, x \neq 0$ .

[6]

c.i.

By using the expression for $f' (x)$ and the result $\sqrt{x^{2}} = |x|$ , show that $f$ is decreasing for $x < 0$ .

[3]

c.ii.

Find an expression for $g^{- 1} (x)$ , justifying your answer.

[5]

State the domain of $g^{- 1}$ .

[1]

Sketch the graph of $y = g^{- 1} (x)$ , clearly indicating any asymptotes with their equations and stating the values of any axes intercepts.

[3]

Markscheme

EITHER

$f (- x) = arcsin (\frac{{(- x)}^{2} - 1}{{(- x)}^{2} + 1}) = arcsin (\frac{x^{2} - 1}{x^{2} + 1}) = f (x)$ R1

a sketch graph of $y = f (x)$ with line symmetry in the $y$ -axis indicated R1

THEN

so $f (x)$ is an even function. AG

[1 mark]

as $x \to \pm \infty, f (x) \to arcsin 1 (\to \frac{π}{2})$ A1

so the horizontal asymptote is $y = \frac{π}{2}$ A1

[2 marks]

attempting to use the quotient rule to find $\frac{d}{d x} (\frac{x^{2} - 1}{x^{2} + 1})$ M1

$\frac{d}{d x} (\frac{x^{2} - 1}{x^{2} + 1}) = \frac{2 x (x^{2} + 1) - 2 x (x^{2} - 1)}{{(x^{2} + 1)}^{2}} (= \frac{4 x}{{(x^{2} + 1)}^{2}})$ A1

attempting to use the chain rule to find $\frac{d}{d x} (arcsin (\frac{x^{2} - 1}{x^{2} + 1}))$ M1

let $u = \frac{x^{2} - 1}{x^{2} + 1}$ and so $y = arcsin u$ and $\frac{d y}{d u} = \frac{1}{\sqrt{1 - u^{2}}}$

$f' (x) = \frac{1}{\sqrt{1 - {(\frac{x^{2} - 1}{x^{2} + 1})}^{2}}} \times \frac{4 x}{{(x^{2} + 1)}^{2}}$ M1

$= \frac{4 x}{\sqrt{{(x^{2} + 1)}^{2} - {(x^{2} - 1)}^{2}}} \times \frac{1}{(x^{2} + 1)}$ A1

$= \frac{4 x}{\sqrt{4 x^{2}}} \times \frac{1}{(x^{2} + 1)}$ A1

$= \frac{2 x}{\sqrt{x^{2}} (x^{2} + 1)}$ AG

[6 marks]

c.i.

$f' (x) = \frac{2 x}{|x| (x^{2} + 1)}$

EITHER

for $x < 0, |x| = - x$ (A1)

so $f' (x) = - \frac{2 x}{x^{2} + 1}$ A1

$|x| > 0$ and $x^{2} + 1 > 0$ A1

$2 x < 0, x < 0$ A1

THEN

$f' (x) < 0$ R1

Note: Award R1 for stating that in $f' (x)$ , the numerator is negative, and the denominator is positive.

so $f$ is decreasing for $x < 0$ AG

Note: Do not accept a graphical solution

[3 marks]

c.ii.

$x = arcsin (\frac{y^{2} - 1}{y^{2} + 1})$ M1

$sin x = \frac{y^{2} - 1}{y^{2} + 1} \Rightarrow y^{2} sin x + sin x = y^{2} - 1$ A1

$y^{2} = \frac{1 + sin x}{1 - sin x}$ A1

domain of $g$ is $x \in ℝ, x \geq 0$ and so the range of $g^{- 1}$ must be $y \in ℝ, y \geq 0$

hence the positive root is taken (or the negative root is rejected) R1

Note: The R1 is dependent on the above A1.

so $(g^{- 1} (x) =) \sqrt{\frac{1 + sin x}{1 - sin x}}$ A1

Note: The final A1 is not dependent on R1 mark.

[5 marks]

domain is $- \frac{π}{2} \leq x < \frac{π}{2}$ A1

Note: Accept correct alternative notations, for example, $⌊ - \frac{π}{2}, \frac{π}{2} ⌊$ or $⌊ - \frac{π}{2}, \frac{π}{2})$ .
Accept $[- 1.57, 1.57 [$ if correct to $3$ s.f.

[1 mark]

A1A1A1

Note: A1 for correct domain and correct range and $y$ -intercept at $y = 1$
A1 for asymptotic behaviour $x \to \frac{π}{2}$
A1 for $x = \frac{π}{2}$
Coordinates are not required.
Do not accept $x = 1.57$ or other inexact values.

[3 marks]

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

21N.2.AHL.TZ0.8a

Question

Consider the curve $C$ given by $y = x - x y \ln (x y)$ where $x > 0, y > 0$ .

Show that $\frac{d y}{d x} + (x \frac{d y}{d x} + y) (1 + \ln (x y)) = 1$ .

[3]

Hence find the equation of the tangent to $C$ at the point where $x = 1$ .

[5]

Markscheme

METHOD 1

attempts to differentiate implicitly including at least one application of the product rule (M1)

$u = x y, v = \ln (x y), \frac{d u}{d x} = x \frac{d y}{d x} + y, \frac{d v}{d x} = (x \frac{d y}{d x} + y) \frac{1}{x y}$

$\frac{d y}{d x} = 1 - [\frac{x y}{x y} (x \frac{d y}{d x} + y) + (x \frac{d y}{d x} + y) \ln (x y)]$ A1

Note: Award (M1)A1 for implicitly differentiating $y = x (1 - y \ln (x y))$ and obtaining $\frac{d y}{d x} = 1 - [\frac{x y}{x y} (x \frac{d y}{d x} + y) + x \frac{d y}{d x} \ln (x y) + y \ln (x y)]$ .

$\frac{d y}{d x} = 1 - [(x \frac{d y}{d x} + y) + (x \frac{d y}{d x} + y) \ln (x y)]$

$\frac{d y}{d x} = 1 - (x \frac{d y}{d x} + y) (1 + \ln (x y))$ A1

$\frac{d y}{d x} + (x \frac{d y}{d x} + y) (1 + \ln (x y)) = 1$ AG

METHOD 2

$y = x - x y \ln x - x y \ln y$

attempts to differentiate implicitly including at least one application of the product rule (M1)

$\frac{d y}{d x} = 1 - (\frac{x y}{x} + (x \frac{d y}{d x} + y) \ln x) - (\frac{x y}{y} \frac{d y}{d x} + (x \frac{d y}{d x} + y) \ln y)$ A1

or equivalent to the above, for example

$\frac{d y}{d x} = 1 - (x \ln x \frac{d y}{d x} + (1 + \ln x) y) - (y \ln y + x (\ln y \frac{d y}{d x} + \frac{d y}{d x}))$

$\frac{d y}{d x} = 1 - x \frac{d y}{d x} (\ln x + \ln y + 1) - y (\ln x + \ln y + 1)$ A1

or equivalent to the above, for example

$\frac{d y}{d x} = 1 - x \frac{d y}{d x} (\ln (x y) + 1) - y (\ln (x y) + 1)$

$\frac{d y}{d x} + (x \frac{d y}{d x} + y) (1 + \ln (x y)) = 1$ AG

METHOD 3

attempt to differentiate implicitly including at least one application of the product rule M1

$u = x \ln (x y), v = y, \frac{d u}{d x} = \ln (x y) + (x \frac{d y}{d x} + y) \frac{x}{x y}, \frac{d v}{d x} = \frac{d y}{d x}$

$\frac{d y}{d x} = 1 - (x \frac{d y}{d x} \ln (x y) + y \ln (x y) + \frac{x y}{x y} (x \frac{d y}{d x} + y))$ A1

$\frac{d y}{d x} = 1 - x \frac{d y}{d x} (\ln (x y) + 1) - y (\ln (x y) + 1)$ A1

$\frac{d y}{d x} + (x \frac{d y}{d x} + y) (1 + \ln (x y)) = 1$ AG

METHOD 4

lets $w = x y$ and attempts to find $\frac{d y}{d x}$ where $y = x - w \ln w$ M1

$\frac{d y}{d x} = 1 - (\frac{d w}{d x} + \frac{d w}{d x} \ln w) (= 1 - \frac{d w}{d x} (1 + \ln w))$ A1

$\frac{d w}{d x} = x \frac{d y}{d x} + y$ A1

$\frac{d y}{d x} = 1 - (x \frac{d y}{d x} + y + (x \frac{d y}{d x} + y) \ln (x y)) (= 1 - (x \frac{d y}{d x} + y) (1 + \ln (x y)))$

$\frac{d y}{d x} + (x \frac{d y}{d x} + y) (1 + \ln (x y)) = 1$ AG

[3 marks]

METHOD 1

substitutes $x = 1$ into $y = x - x y \ln (x y)$ (M1)

$y = 1 - y \ln y \Rightarrow y = 1$ A1

substitutes $x = 1$ and their non-zero value of $y$ into $\frac{d y}{d x} + (x \frac{d y}{d x} + y) (1 + \ln (x y)) = 1$ (M1)

$2 \frac{d y}{d x} = 0 (\frac{d y}{d x} = 0)$ A1

equation of the tangent is $y = 1$ A1

METHOD 2

substitutes $x = 1$ into $\frac{d y}{d x} + (x \frac{d y}{d x} + y) (1 + \ln (x y)) = 1$ (M1)

$\frac{d y}{d x} + (\frac{d y}{d x} + y) (1 + \ln (y)) = 1$

EITHER

correctly substitutes $\ln y = \frac{1 - y}{y}$ into $\frac{d y}{d x} + (\frac{d y}{d x} + y) (1 + \ln (x y)) = 1$ A1

$\frac{d y}{d x} (1 + \frac{1}{y}) = 0 \Rightarrow \frac{d y}{d x} = 0 (y = 1)$ A1

correctly substitutes $y + y \ln y = 1$ into $\frac{d y}{d x} + (\frac{d y}{d x} + y) (1 + \ln (x y)) = 1$ A1

$\frac{d y}{d x} (2 + \ln y) = 0 \Rightarrow \frac{d y}{d x} = 0 (y = 1)$ A1

THEN

substitutes $x = 1$ into $y = x - x y \ln (x y)$ (M1)

$y = 1 - y \ln y \Rightarrow y = 1$

equation of the tangent is $y = 1$ A1

[5 marks]

Examiners report

[N/A]

21N.2.AHL.TZ0.10b

Question

Consider the function $f (x) = \frac{x^{2} - x - 12}{2 x - 15}, x \in ℝ, x \neq \frac{15}{2}$ .

Find the coordinates where the graph of $f$ crosses the

$x$ -axis.

[2]

a.i.

$y$ -axis.

[1]

a.ii.

Write down the equation of the vertical asymptote of the graph of $f$ .

[1]

The oblique asymptote of the graph of $f$ can be written as $y = a x + b$ where $a, b \in ℚ$ .

Find the value of $a$ and the value of $b$ .

[4]

Sketch the graph of $f$ for $- 30 \leq x \leq 30$ , clearly indicating the points of intersection with each axis and any asymptotes.

[3]

Express $\frac{1}{f (x)}$ in partial fractions.

[3]

e.i.

Hence find the exact value of $\int_{0}^{3} \frac{1}{f (x)} d x$ , expressing your answer as a single logarithm.

[4]

e.ii.

Markscheme

Note: In part (a), penalise once only, if correct values are given instead of correct coordinates.

attempts to solve $x^{2} - x - 12 = 0$ (M1)

$(- 3, 0)$ and $(4, 0)$ A1

[2 marks]

a.i.

Note: In part (a), penalise once only, if correct values are given instead of correct coordinates.

$(0, \frac{4}{5})$ A1

[1 mark]

a.ii.

$x = \frac{15}{2}$ A1

Note: Award A0 for $x \neq \frac{15}{2}$ .
Award A1 in part (b), if $x = \frac{15}{2}$ is seen on their graph in part (d).

[1 mark]

METHOD 1

$(a x + b) (2 x - 15) \equiv x^{2} - x - 12$

attempts to expand $(a x + b) (2 x - 15)$ (M1)

$2 a x^{2} - 15 a x + 2 b x - 15 b \equiv x^{2} - x - 12$

$a = \frac{1}{2}$ A1

equates coefficients of $x$ (M1)

$- 1 = - \frac{15}{2} + 2 b$

$b = \frac{13}{4}$ A1

$(y = \frac{x}{2} + \frac{13}{4})$

METHOD 2

attempts division on $\frac{x^{2} - x - 12}{2 x - 15}$ M1

$\frac{x}{2} + \frac{13}{4} + \dots$ M1

$a = \frac{1}{2}$ A1

$b = \frac{13}{4}$ A1

$(y = \frac{x}{2} + \frac{13}{4})$

METHOD 3

$a = \frac{1}{2}$ A1

$\frac{x^{2} - x - 12}{2 x - 15} \equiv \frac{x}{2} + b + \frac{c}{2 x - 15}$ M1

$x^{2} - x - 12 \equiv \frac{(2 x - 15) x}{2} + (2 x - 15) b + c$

equates coefficients of $x$ : (M1)

$- 1 = - \frac{15}{2} + 2 b$

$b = \frac{13}{4}$ A1

$(y = \frac{x}{2} + \frac{13}{4})$

METHOD 4

attempts division on $\frac{x^{2} - x - 12}{2 x - 15}$ M1

$\frac{x^{2} - x - 12}{2 x - 15} = \frac{x}{2} + \frac{\frac{13 x}{2} - 12}{2 x - 15}$

$a = \frac{1}{2}$ A1

$\frac{\frac{13 x}{2} - 12}{2 x - 15} = \frac{13}{4} + \dots$ M1

$b = \frac{13}{4}$ A1

$(y = \frac{x}{2} + \frac{13}{4})$

[4 marks]

two branches with approximately correct shape (for $- 30 \leq x \leq 30$ ) A1

their vertical and oblique asymptotes in approximately correct positions with both branches showing correct asymptotic behaviour to these asymptotes A1

their axes intercepts in approximately the correct positions A1

Note: Points of intersection with the axes and the equations of asymptotes are not required to be labelled.

[3 marks]

attempts to split into partial fractions: (M1)

$\frac{2 x - 15}{(x + 3) (x - 4)} \equiv \frac{A}{x + 3} + \frac{B}{x - 4}$

$2 x - 15 \equiv A (x - 4) + B (x + 3)$

$A = 3$ A1

$B = - 1$ A1

$(\frac{3}{x + 3} - \frac{1}{x - 4})$

[3 marks]

e.i.

$\int_{0}^{3} (\frac{3}{x + 3} - \frac{1}{x - 4}) d x$

attempts to integrate and obtains two terms involving ‘ln’ (M1)

$= {[3 \ln |x + 3| - \ln |x - 4|]}_{0}^{3}$ A1

$= 3 \ln 6 - \ln 1 - 3 \ln 3 + \ln 4$ A1

$= 3 \ln 2 + \ln 4 (= \ln 8 + \ln 4)$

$= \ln 32 (= 5 \ln 2)$ A1

Note: The final A1 is dependent on the previous two A marks.

[4 marks]

e.ii.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

e.i.

[N/A]

e.ii.

22M.2.AHL.TZ2.6

Question

The following diagram shows the curve $\frac{x^{2}}{36} + \frac{{(y - 4)}^{2}}{16} = 1$ , where $h \leq y \leq 4$ .

The curve from point $Q$ to point $B$ is rotated $360 °$ about the $y$ -axis to form the interior surface of a bowl. The rectangle $OPQR$ , of height $h cm$ , is rotated $360 °$ about the $y$ -axis to form a solid base.

The bowl is assumed to have negligible thickness.

Given that the interior volume of the bowl is to be $285 {cm}^{3}$ , determine the height of the base.

Markscheme

attempts to express $x^{2}$ in terms of $y$ (M1)

$V = π \int_{h}^{4} 36 (1 - \frac{{(y - 4)}^{2}}{16}) d y$ A1

Note: Correct limits are required.

Attempts to solve $π \int_{h}^{4} 36 (1 - \frac{{(y - 4)}^{2}}{16}) d y = 285$ for $h$ (M1)

Note: Award M1 for attempting to solve $36 π (\frac{h^{3}}{48} - \frac{h^{2}}{4} + \frac{8}{3}) = 285$ or equivalent for $h$ .

$h = 0.7926 \dots$

$h = 0.793$ (cm) A2

[5 marks]

Examiners report

This question was a struggle for many candidates. To start with, many candidates had difficulty understanding the diagram. Some candidates tried to include the base in their equation.

Because of this confusion, the question was poorly attempted. Some only received one mark for rearranging the equation to make $x^{2}$ the subject but were unable to set the correct definite integral with correct terminals. Again, many candidates tried to solve by hand instead of using their GDC. The correct answer was not seen that often.

Those candidates who recognised that the volume was around the y -axis and used their GDC to solve, usually achieved full marks for this question.

22M.2.AHL.TZ2.7a

Question

Consider $\lim_{x \to 0} \frac{arctan (\cos x) - k}{x^{2}}$ , where $k \in ℝ$ .

Show that a finite limit only exists for $k = \frac{π}{4}$ .

[2]

Using l’Hôpital’s rule, show algebraically that the value of the limit is $- \frac{1}{4}$ .

[6]

Markscheme

(as $\lim_{x \to 0} x^{2} = 0$ , the indeterminate form $\frac{0}{0}$ is required for the limit to exist)

$\Rightarrow \lim_{x \to 0} (arctan (\cos x) - k) = 0$ M1

$arctan 1 - k = 0 (k = arctan 1)$ A1

so $k = \frac{π}{4}$ AG

Note: Award M1A0 for using $k = \frac{π}{4}$ to show the limit is $\frac{0}{0}$ .

[2 marks]

$\lim_{x \to 0} \frac{arctan (\cos x) - \frac{π}{4}}{x^{2}} (= \frac{0}{0})$

$= \lim_{x \to 0} \frac{\frac{- \sin x}{1 + \cos^{2} x}}{2 x}$ A1A1

Note: Award A1 for a correct numerator and A1 for a correct denominator.

recognises to apply l’Hôpital’s rule again (M1)

$= \lim_{x \to 0} \frac{\frac{- \sin x}{1 + \cos^{2} x}}{2 x} (= \frac{0}{0})$

Note: Award M0 if their limit is not the indeterminate form $\frac{0}{0}$ .

EITHER

$= \lim_{x \to 0} \frac{\frac{- \cos x (1 + \cos^{2} x) - 2 \sin^{2} x \cos x}{{(1 + \cos^{2} x)}^{2}}}{2}$ A1A1

Note: Award A1 for a correct first term in the numerator and A1 for a correct second term in the numerator.

$\lim_{x \to 0} \frac{- \cos x}{2 (1 + \cos^{2} x) - 4 x \sin x \cos x}$ A1A1

Note: Award A1 for a correct numerator and A1 for a correct denominator.

THEN

substitutes $x = 0$ into the correct expression to evaluate the limit A1

Note: The final A1 is dependent on all previous marks.

$= - \frac{1}{4}$ AG

[6 marks]

Examiners report

Part (a) Many candidates recognised the indeterminate form and provided a nice algebraic proof. Some verified by substituting the given value. Therefore, there is a need to teach the candidates the difference between proof and verification. Only a few candidates were able to give a complete 'show that' proof.

Part (b) Many candidates realised that they needed to apply the L'Hôpital's rule twice. There were many mistakes in differentiation using the chain rule. Not all candidates clearly showed the final substitution.

[N/A]

22M.2.AHL.TZ2.10d

Question

A scientist conducted a nine-week experiment on two plants, $A$ and $B$ , of the same species. He wanted to determine the effect of using a new plant fertilizer. Plant $A$ was given fertilizer regularly, while Plant $B$ was not.

The scientist found that the height of Plant $A, h_{A} cm$ , at time $t$ weeks can be modelled by the function $h_{A} (t) = \sin (2 t + 6) + 9 t + 27$ , where $0 \leq t \leq 9$ .

The scientist found that the height of Plant $B, h_{B} cm$ , at time $t$ weeks can be modelled by the function $h_{B} (t) = 8 t + 32$ , where $0 \leq t \leq 9$ .

Use the scientist’s models to find the initial height of

Plant $B$ .

[1]

a.i.

Plant $A$ correct to three significant figures.

[2]

a.ii.

Find the values of $t$ when $h_{A} (t) = h_{B} (t)$ .

[3]

For $t > 6$ , prove that Plant $A$ was always taller than Plant $B$ .

[3]

For $0 \leq t \leq 9$ , find the total amount of time when the rate of growth of Plant $B$ was greater than the rate of growth of Plant $A$ .

[6]

Markscheme

$32$ (cm) A1

[1 mark]

a.i.

$h_{A} (0) = \sin (6) + 27$ (M1)

$= 26.7205 \dots$

$= 26.7$ (cm) A1

[2 marks]

a.ii.

attempts to solve $h_{A} (t) = h_{B} (t)$ for $t$ (M1)

$t = 4.0074 \dots, 4.7034 \dots, 5.88332 \dots$

$t = 4.01, 4.70, 5.88$ (weeks) A2

[3 marks]

$h_{A} (t) - h_{B} (t) = \sin (2 t + 6) + t - 5$ A1

EITHER

for $t > 6, t - 5 > 1$ A1

and as $\sin (2 t + 6) \geq - 1 \Rightarrow h_{A} (t) - h_{B} (t) > 0$ R1

the minimum value of $\sin (2 t + 6) = - 1$ R1

so for $t > 6, h_{A} (t) - h_{B} (t) = t - 6 > 0$ A1

THEN

hence for $t > 6$ , Plant $A$ was always taller than Plant $B$ AG

[3 marks]

recognises that $h_{A}' (t)$ and $h_{B}' (t)$ are required (M1)

attempts to solve $h_{A}' (t) = h_{B}' (t)$ for $t$ (M1)

$t = 1.18879 \dots$ and $2.23598 \dots$ OR $4.33038 \dots$ and $5.37758 \dots$ OR $7.47197 \dots$ and $8.51917 \dots$ (A1)

Note: Award full marks for $t = \frac{4 π}{3} - 3, \frac{5 π}{3} - 3, (\frac{7 π}{3} - 3, \frac{8 π}{3} - 3 \frac{10 π}{3} - 3, \frac{11 π}{3} - 3)$ .

Award subsequent marks for correct use of these exact values.

$1.18879 \dots < t < 2.23598 \dots$ OR $4.33038 \dots < t < 5.37758 \dots$ OR $7.47197 \dots < t < 8.51917 \dots$ (A1)

attempts to calculate the total amount of time (M1)

$3 (2.2359 \dots - 1.1887 \dots) (= 3 ((\frac{5 π}{3} - 3) - (\frac{4 π}{3} - 3)))$

$= 3.14 (= π)$ (weeks) A1

[6 marks]

Examiners report

Part (a) In general, very well done, most students scored full marks. Some though had an incorrect answer for part(a)(ii) because they had their GDC in degrees.

Part (b) Well attempted. Some accuracy errors and not all candidates listed all three values.

Part (c) Most students tried a graphical approach (but this would only get them one out of three marks) and only some provided a convincing algebraic justification. Many candidates tried to explain in words without a convincing mathematical justification or used numerical calculations with specific time values. Some arrived at the correct simplified equation for the difference in heights but could not do much with it. Then only a few provided a correct mathematical proof.

Part (d) In general, well attempted by many candidates. The common error was giving the answer as 3.15 due to the pre-mature rounding. Some candidates only provided the values of time when the rates are equal, some intervals rather than the total time.

a.i.

[N/A]

a.ii.

[N/A]

22M.2.AHL.TZ1.10d

Question

Consider the function $f (x) = \sqrt{x^{2} - 1}$ , where $1 \leq x \leq 2$ .

The curve $y = f (x)$ is rotated $2 π$ about the $y$ -axis to form a solid of revolution that is used to model a water container.

At $t = 0$ , the container is empty. Water is then added to the container at a constant rate of $0.4 m^{3} s^{- 1}$ .

Sketch the curve $y = f (x)$ , clearly indicating the coordinates of the endpoints.

[2]

Show that the inverse function of $f$ is given by $f^{- 1} (x) = \sqrt{x^{2} + 1}$ .

[3]

b.i.

State the domain and range of $f^{- 1}$ .

[2]

b.ii.

Show that the volume, $V m^{3}$ , of water in the container when it is filled to a height of $h$ metres is given by $V = π (\frac{1}{3} h^{3} + h)$ .

[3]

c.i.

Hence, determine the maximum volume of the container.

[2]

c.ii.

Find the time it takes to fill the container to its maximum volume.

[2]

Find the rate of change of the height of the water when the container is filled to half its maximum volume.

[6]

Markscheme

correct shape (concave down) within the given domain $1 \leq x \leq 2$ A1

$(1, 0)$ and $(2, \sqrt{3}) (= (2, 1.73))$ A1

Note: The coordinates of endpoints may be seen on the graph or marked on the axes.

[2 marks]

interchanging $x$ and $y$ (seen anywhere) M1

$x = \sqrt{y^{2} - 1}$

$x^{2} = y^{2} - 1$ A1

$y = \sqrt{x^{2} + 1}$ A1

$f^{- 1} (x) = \sqrt{x^{2} + 1}$ AG

[3 marks]

b.i.

$0 \leq x \leq \sqrt{3}$ OR domain $[0, \sqrt{3}] (= [0, 1.73])$ A1

$1 \leq y \leq 2$ OR $1 \leq f^{- 1} (x) \leq 2$ OR range $[1, 2]$ A1

[2 marks]

b.ii.

attempt to substitute $x = \sqrt{y^{2} + 1}$ into the correct volume formula (M1)

$V = π \int_{0}^{h} {(\sqrt{y^{2} + 1})}^{2} d y (= π \int_{0}^{h} (y^{2} + 1) d y)$ A1

$= π {[\frac{1}{3} y^{3} + y]}_{0}^{h}$ A1

$= π (\frac{1}{3} h^{3} + h)$ AG

Note: Award marks as appropriate for correct work using a different variable e.g. $π \int_{0}^{h} {(\sqrt{x^{2} + 1})}^{2} d x$

[3 marks]

c.i.

attempt to substitute $h = \sqrt{3} (= 1.732 \dots)$ into $V$ (M1)

$V = 10.8828 \dots$

$V = 10.9 (m^{3}) (= 2 \sqrt{3} π) (m^{3})$ A1

[2 marks]

c.ii.

time $= \frac{10.8828 \dots}{0.4} (= \frac{2 \sqrt{3} π}{0.4})$ (M1)

$= 27.207 \dots$

$= 27.2 (= 5 \sqrt{3} π) (s)$ A1

[2 marks]

attempt to find the height of the tank when $V = 5.4414 \dots (= \sqrt{3} π)$ (M1)

$π (\frac{1}{3} h^{3} + h) = 5.4414 \dots (= \sqrt{3} π)$

$h = 1.1818 \dots$ (A1)

attempt to use the chain rule or differentiate $V = π (\frac{1}{3} h^{3} + h)$ with respect to $t$ (M1)

$\frac{d h}{d t} = \frac{d h}{d V} \times \frac{d V}{d t} = \frac{1}{π (h^{2} + 1)} \times \frac{d V}{d t}$ OR $\frac{d V}{d t} = π (h^{2} + 1) \frac{d h}{d t}$ (A1)

attempt to substitute their $h$ and $\frac{d V}{d t} = 0.4$ (M1)

$\frac{d h}{d t} = \frac{0.4}{π (1.1818 \dots^{2} + 1)} = 0.053124 \dots$

$= 0.0531 ({m s}^{- 1})$ A1

[6 marks]

Examiners report

Part a) was generally well done, with the most common errors being to use an incorrect domain or not to give the coordinates of the endpoints. Some graphs appeared to be straight lines; some candidates drew sketches which were too small which made it more difficult for them to show the curvature.

Most candidates were able to show the steps to find an inverse function in part b), although occasionally a candidate did not explicitly swop the x and y variables before writing down the inverse function, which was given in the question. Many candidates struggled to identify the domain and range of the inverse, despite having a correct graph.

Part c) required a rotation around the y-axis, but a number of candidates attempted to rotate around the x-axis or failed to include limits. In the same vein, many substituted 2 into the formula instead of the square root of 3 when answering the second part. Many subsequently gained follow through marks on part d).

There were a number of good attempts at related rates in part e), with the majority differentiating V with respect to t, using implicit differentiation. However, many did not find the value of h which corresponded to halving the volume, and a number did not differentiate with respect to t, only with respect to h.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

22M.2.AHL.TZ2.11e

Question

Two airplanes, $A$ and $B$ , have position vectors with respect to an origin $O$ given respectively by

$r_{A} = (\begin{matrix} 19 \\ - 1 \\ 1 \end{matrix}) + t (\begin{matrix} - 6 \\ 2 \\ 4 \end{matrix})$

$r_{B} = (\begin{matrix} 1 \\ 0 \\ 12 \end{matrix}) + t (\begin{matrix} 4 \\ 2 \\ - 2 \end{matrix})$

where $t$ represents the time in minutes and $0 \leq t \leq 2.5$ .

Entries in each column vector give the displacement east of $O$ , the displacement north of $O$ and the distance above sea level, all measured in kilometres.

The two airplanes’ lines of flight cross at point $P$ .

Find the three-figure bearing on which airplane $B$ is travelling.

[2]

Show that airplane $A$ travels at a greater speed than airplane $B$ .

[2]

Find the acute angle between the two airplanes’ lines of flight. Give your answer in degrees.

[4]

Find the coordinates of $P$ .

[5]

d.i.

Determine the length of time between the first airplane arriving at $P$ and the second airplane arriving at $P$ .

[2]

d.ii.

Let $D (t)$ represent the distance between airplane $A$ and airplane $B$ for $0 \leq t \leq 2.5$ .

Find the minimum value of $D (t)$ .

[5]

Markscheme

let $ϕ$ be the required angle (bearing)

EITHER

$ϕ = 90 ° - arctan \frac{1}{2} (= arctan 2)$ (M1)

Note: Award M1 for a labelled sketch.

$\cos ϕ = \frac{(\begin{matrix} 0 \\ 1 \end{matrix}) \cdot (\begin{matrix} 4 \\ 2 \end{matrix})}{\sqrt{1} \times \sqrt{20}} (= 0.4472 \dots, = \frac{1}{\sqrt{5}})$ (M1)

$ϕ = arccos (0.4472 \dots)$

THEN

$063 °$ A1

Note: Do not accept $063.6 °$ or $63.4 °$ or $1 . 10^{c}$ .

[2 marks]

METHOD 1

let $|b_{A}|$ be the speed of $A$ and let $|b_{B}|$ be the speed of $B$

attempts to find the speed of one of $A$ or $B$ (M1)

$|b_{A}| = \sqrt{{(- 6)}^{2} + 2^{2} + 4^{2}}$ or $|b_{B}| = \sqrt{4^{2} + 2^{2} + {(- 2)}^{2}}$

Note: Award M0 for $|b_{A}| = \sqrt{19^{2} + {(- 1)}^{2} + 1^{2}}$ and $|b_{B}| = \sqrt{1^{2} + 0^{2} + 12^{2}}$ .

$|b_{A}| = 7.48 \dots (= \sqrt{56})$ (km min^-1) and $|b_{B}| = 4.89 \dots (= \sqrt{24})$ (km min^-1) A1

$|b_{A}| > |b_{B}|$ so $A$ travels at a greater speed than $B$ AG

METHOD 2

attempts to use $speed = \frac{distance}{time}$

${speed}_{A} = \frac{|r_{A} (t_{2}) - r_{A} (t_{1})|}{t_{2} - t_{1}}$ and ${speed}_{B} = \frac{|r_{B} (t_{2}) - r_{B} (t_{1})|}{t_{2} - t_{1}}$ (M1)

for example:

${speed}_{A} = \frac{|r_{A} (1) - r_{A} (0)|}{1}$ and ${speed}_{B} = \frac{|r_{B} (1) - r_{B} (0)|}{1}$

${speed}_{A} = \frac{\sqrt{{(- 6)}^{2} + 2^{2} + 4^{2}}}{1}$ and ${speed}_{B} = \frac{\sqrt{4^{2} + 2^{2} + 2^{2}}}{1}$

${speed}_{A} = 7.48 \dots (2 \sqrt{14})$ and ${speed}_{B} = 4.89 \dots (\sqrt{24})$ A1

${speed}_{A} > {speed}_{B}$ so $A$ travels at a greater speed than $B$ AG

[2 marks]

attempts to use the angle between two direction vectors formula (M1)

$\cos θ = \frac{(- 6) (4) + (2) (2) + (4) (- 2)}{\sqrt{{(- 6)}^{2} + 2^{2} + 4^{2}} \sqrt{4^{2} + 2^{2} + {(- 2)}^{2}}}$ (A1)

$\cos θ = - 0.7637 \dots (= - \frac{7}{\sqrt{84}})$ or $θ = arccos (- 0.7637 \dots) (= 2.4399 \dots)$

attempts to find the acute angle $180 ° - θ$ using their value of $θ$ (M1)

$= 40.2 °$ A1

[4 marks]

for example, sets $r_{A} (t_{1}) = r_{B} (t_{2})$ and forms at least two equations (M1)

$19 - 6 t_{1} = 1 + 4 t_{2}$

$- 1 + 2 t_{1} = 2 t_{2}$

$1 + 4 t_{1} = 12 - 2 t_{2}$

Note: Award M0 for equations involving $t$ only.

EITHER

attempts to solve the system of equations for one of $t_{1}$ or $t_{2}$ (M1)

$t_{1} = 2$ or $t_{2} = \frac{3}{2}$ A1

attempts to solve the system of equations for $t_{1}$ and $t_{2}$ (M1)

$t_{1} = 2$ or $t_{2} = \frac{3}{2}$ A1

THEN

substitutes their $t_{1}$ or $t_{2}$ value into the corresponding $r_{A}$ or $r_{B}$ (M1)

$P (7, 3, 9)$ A1

Note: Accept $\vec{OP} = (\begin{matrix} 7 \\ 3 \\ 9 \end{matrix})$ . Accept $7$ km east of $O$ , $3$ km north of $O$ and $9$ km above sea level.

[5 marks]

d.i.

attempts to find the value of $t_{1} - t_{2}$ (M1)

$t_{1} - t_{2} = 2 - \frac{3}{2}$

$0.5$ minutes ( $30$ seconds) A1

[2 marks]

d.ii.

EITHER

attempts to find $r_{B} - r_{A}$ (M1)

$r_{B} - r_{A} = (\begin{matrix} - 18 \\ 1 \\ 11 \end{matrix}) + t (\begin{matrix} 10 \\ 0 \\ - 6 \end{matrix})$

attempts to find their $D (t)$ (M1)

$D (t) = \sqrt{{(10 t - 18)}^{2} + 1 + {(11 - 6 t)}^{2}}$ A1

attempts to find $r_{A} - r_{B}$ (M1)

$r_{A} - r_{B} = (\begin{matrix} 18 \\ - 1 \\ - 11 \end{matrix}) + t (\begin{matrix} - 10 \\ 0 \\ 6 \end{matrix})$

attempts to find their $D (t)$ (M1)

$D (t) = \sqrt{{(18 - 10 t)}^{2} + {(- 1)}^{2} + {(- 11 + 6 t)}^{2}}$ A1

Note: Award M0M0A0 for expressions using two different time parameters.

THEN

either attempts to find the local minimum point of $D (t)$ or attempts to find the value of $t$ such that $D' (t) = 0$ (or equivalent) (M1)

$t = 1.8088 \dots (= \frac{123}{68})$

$D (t) = 1.01459 \dots$

minimum value of $D (t)$ is $1.01 (= \frac{\sqrt{1190}}{34})$ (km) A1

Note: Award M0 for attempts at the shortest distance between two lines.

[5 marks]

Examiners report

General comment about this question: many candidates were not exposed to this setting of vectors question and were rather lost.

Part (a) Probably the least answered question on the whole paper. Many candidates left it blank, others tried using 3D vectors. Out of those who calculated the angle correctly, only a small percentage were able to provide the correct true bearing as a 3-digit figure.

Part (b) Well done by many candidates who used the direction vectors to calculate and compare the speeds. A number of candidates tried to use the average rate of change but mostly unsuccessfully.

Part (c) Most candidates used the correct vectors and the formula to obtain the obtuse angle. Then only some read the question properly to give the acute angle in degrees, as requested.

Part (d) Well done by many candidates who used two different parameters. They were able to solve and obtain two values for time, the difference in minutes and the correct point of intersection. A number of candidates only had one parameter, thus scoring no marks in part (d) (i). The frequent error in part (d)(ii) was providing incorrect units.

Part (e) Many correct answers were seen with an efficient way of setting the question and using their GDC to obtain the answer, graphically or numerically. Some gave time only instead of actually giving the minimal distance. A number of candidates tried to find the distance between two skew lines ignoring the fact that the lines intersect.

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

22M.2.AHL.TZ1.12a

Question

Consider the differential equation $x^{2} \frac{d y}{d x} = y^{2} - 2 x^{2}$ for $x > 0$ and $y > 2 x$ . It is given that $y = 3$ when $x = 1$ .

Use Euler’s method, with a step length of $0.1$ , to find an approximate value of $y$ when $x = 1.5$ .

[4]

Use the substitution $y = v x$ to show that $x \frac{d v}{d x} = v^{2} - v - 2$ .

[3]

By solving the differential equation, show that $y = \frac{8 x + x^{4}}{4 - x^{3}}$ .

[10]

c.i.

Find the actual value of $y$ when $x = 1.5$ .

[1]

c.ii.

Using the graph of $y = \frac{8 x + x^{4}}{4 - x^{3}}$ , suggest a reason why the approximation given by Euler’s method in part (a) is not a good estimate to the actual value of $y$ at $x = 1.5$ .

[1]

c.iii.

Markscheme

attempt to use Euler’s method (M1)

$x_{n + 1} = x_{n} + 0.1; y_{n + 1} = y_{n} + 0.1 \times \frac{d y}{d x}$ , where $\frac{d y}{d x} = \frac{y^{2} - 2 x^{2}}{x^{2}}$

correct intermediate $y$ -values (A1)(A1)

$3.7, 4.63140 \dots, 5.92098, 7.79542 \dots$

Note: A1 for any two correct $y$ -values seen

$y = 10.6958 \dots$

$y = 10.7$ A1

Note: For the final A1, the value $10.7$ must be the last value in a table or a list, or be given as a final answer, not just embedded in a table which has further lines.

[4 marks]

$y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$ (A1)

replacing $y$ with $v x$ and $\frac{d y}{d x}$ with $v + x \frac{d v}{d x}$ M1

$x^{2} \frac{d y}{d x} = y^{2} - 2 x^{2} \Rightarrow x^{2} (v + x \frac{d v}{d x}) = v^{2} x^{2} - 2 x^{2}$ A1

$v + x \frac{d v}{d x} = v^{2} - 2$ (since $x > 0$ )

$x \frac{d v}{d x} = v^{2} - v - 2$ AG

[3 marks]

attempt to separate variables $v$ and $x$ (M1)

$\int \frac{d v}{v^{2} - v - 2} = \int \frac{d x}{x}$

$\int \frac{d v}{(v - 2) (v + 1)} = \int \frac{d x}{x}$ (A1)

attempt to express in partial fraction form M1

$\frac{1}{(v - 2) (v + 1)} \equiv \frac{A}{v - 2} + \frac{B}{v + 1}$

$\frac{1}{(v - 2) (v + 1)} = \frac{1}{3} (\frac{1}{v - 2} - \frac{1}{v + 1})$ A1

$\frac{1}{3} \int (\frac{1}{v - 2} - \frac{1}{v + 1}) d v = \int \frac{d x}{x}$

$\frac{1}{3} (\ln |v - 2| - \ln |v + 1|) = \ln |x| (+ c)$ A1

Note: Condone absence of modulus signs throughout.

EITHER

attempt to find $c$ using $x = 1, y = 3, v = 3$ M1

$c = \frac{1}{3} \ln \frac{1}{4}$

$\frac{1}{3} (\ln |v - 2| - \ln |v + 1|) = \ln |x| + \frac{1}{3} \ln \frac{1}{4}$

expressing both sides as a single logarithm (M1)

$\ln |\frac{v - 2}{v + 1}| = \ln (\frac{{|x|}^{3}}{4})$

expressing both sides as a single logarithm (M1)

$\ln |\frac{v - 2}{v + 1}| = \ln (A {|x|}^{3})$

attempt to find $A$ using $x = 1, y = 3, v = 3$ M1

$A = \frac{1}{4}$

THEN

$|\frac{v - 2}{v + 1}| = \frac{1}{4} x^{3}$ (since $x > 0$ )

substitute $v = \frac{y}{x}$ (seen anywhere) M1

$\frac{\frac{y}{x} - 2}{\frac{y}{x} + 1} = \frac{1}{4} x^{3}$ (since $y > 2 x$ )

$(\Rightarrow \frac{y - 2 x}{y + x} = \frac{1}{4} x^{3})$

attempt to make $y$ the subject M1

$y - \frac{x^{3} y}{4} = 2 x + \frac{x^{4}}{4}$ A1

$y = \frac{8 x + x^{4}}{4 - x^{3}}$ AG

[10 marks]

c.i.

actual value at $y (1.5) = 27.3$ A1

[1 mark]

c.ii.

gradient changes rapidly (during the interval considered) OR

the curve has a vertical asymptote at $x = \sqrt[3]{4} (= 1.5874 \dots)$ R1

[1 mark]

c.iii.

Examiners report

Most candidates showed evidence of an attempt to use Euler's method in part a), although very few explicitly wrote down the formulae, they used in order to calculate successive y-value. In addition, many seemed to take a step-by-step approach rather than using the recursive capabilities of the graphical display calculator.

There were many good attempts at part b), but not all candidates recognised that this would help them to solve part c).

Part c) was done very well by many candidates, although there were a significant number who failed to recognise the need for partial fractions and could not make further progress. A common error was to integrate without a constant of integration, which meant that the initial condition could not be used. The reasoning given for the estimate being poor was often too vague and did not address the specific nature of the function given clearly enough.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

c.iii.

22M.2.AHL.TZ2.12a

Question

The population, $P$ , of a particular species of marsupial on a small remote island can be modelled by the logistic differential equation

$\frac{d P}{d t} = k P (1 - \frac{P}{N})$

where $t$ is the time measured in years and $k, N$ are positive constants.

The constant $N$ represents the maximum population of this species of marsupial that the island can sustain indefinitely.

Let $P_{0}$ be the initial population of marsupials.

In the context of the population model, interpret the meaning of $\frac{d P}{d t}$ .

[1]

Show that $\frac{d^{2} P}{d t^{2}} = k^{2} P (1 - \frac{P}{N}) (1 - \frac{2 P}{N})$ .

[4]

Hence show that the population of marsupials will increase at its maximum rate when $P = \frac{N}{2}$ . Justify your answer.

[5]

Hence determine the maximum value of $\frac{d P}{d t}$ in terms of $k$ and $N$ .

[2]

By solving the logistic differential equation, show that its solution can be expressed in the form

$k t = \ln \frac{P}{P_{0}} (\frac{N - P_{0}}{N - P})$ .

[7]

After $10$ years, the population of marsupials is $3 P_{0}$ . It is known that $N = 4 P_{0}$ .

Find the value of $k$ for this population model.

[2]

Markscheme

rate of growth (change) of the (marsupial) population (with respect to time) A1

[1 mark]

Note: Do not accept growth (change) in the (marsupials) population per year.

METHOD 1

attempts implicit differentiation on $\frac{d P}{d t} = k P - \frac{k P^{2}}{N}$ be expanding $k P (1 - \frac{P}{N})$ (M1)

$\frac{d^{2} P}{d t^{2}} = k \frac{d P}{d t} - 2 \frac{k P}{N} \frac{d P}{d t}$ A1A1

$= k \frac{d P}{d t} (1 - \frac{2 P}{N})$ A1

$\frac{d P}{d t} = k P (1 - \frac{P}{N})$ and so $\frac{d^{2} P}{d t^{2}} = k^{2} P (1 - \frac{P}{N}) (1 - \frac{2 P}{N})$ AG

METHOD 2

attempts implicit differentiation (product rule) on $\frac{d P}{d t} = k P (1 - \frac{P}{N})$ M1

$\frac{d^{2} P}{d t^{2}} = k \frac{d P}{d t} (1 - \frac{P}{N}) + k P (- (\frac{1}{N}) \frac{d P}{d t})$ A1

substitutes $\frac{d P}{d t} = k P (1 - \frac{P}{N})$ into their $\frac{d^{2} P}{d t^{2}}$ M1

$\frac{d^{2} P}{d t^{2}} = k (k P (1 - \frac{P}{N})) (1 - \frac{P}{N}) + k P (- (\frac{1}{N}) k P (1 - \frac{P}{N}))$

$= k^{2} P {(1 - \frac{P}{N})}^{2} - k^{2} P (1 - \frac{P}{N}) (\frac{P}{N})$

$= k^{2} P (1 - \frac{P}{N}) (1 - \frac{P}{N} - \frac{P}{N})$ A1

so $\frac{d^{2} P}{d t^{2}} = k^{2} P (1 - \frac{P}{N}) (1 - \frac{2 P}{N})$ AG

[4 marks]

$\frac{d^{2} P}{d t^{2}} = 0 \Rightarrow k^{2} P (1 - \frac{P}{N}) (1 - \frac{2 P}{N}) = 0$ (M1)

$P = 0, \frac{N}{2}, N$ A2

Note: Award A1 for $P = \frac{N}{2}$ only.

uses the second derivative to show that concavity changes at $P = \frac{N}{2}$ or the first derivative to show a local maximum at $P = \frac{N}{2}$ M1

EITHER

a clearly labelled correct sketch of $\frac{d^{2} P}{d t^{2}}$ versus $P$ showing $P = \frac{N}{2}$ corresponding to a local maximum point for $\frac{d P}{d t}$ R1

a correct and clearly labelled sign diagram (table) showing $P = \frac{N}{2}$ corresponding to a local maximum point for $\frac{d P}{d t}$ R1

for example, $\frac{d^{2} P}{d t^{2}} = \frac{3 k^{2} N}{32} (> 0)$ with $P = \frac{N}{4}$ and $\frac{d^{2} P}{d t^{2}} = \frac{3 k^{2} N}{32} (< 0)$ with $P = \frac{3 N}{4}$ showing $P = \frac{N}{2}$ corresponds to a local maximum point for $\frac{d P}{d t}$ R1

so the population is increasing at its maximum rate when $P = \frac{N}{2}$ AG

[5 marks]

substitutes $P = \frac{N}{2}$ into $\frac{d P}{d t}$ (M1)

$\frac{d P}{d t} = k (\frac{N}{2}) (1 - \frac{\frac{N}{2}}{N})$

the maximum value of $\frac{d P}{d t}$ is $\frac{k N}{4}$ A1

[2 marks]

METHOD 1

attempts to separate variables M1

$\int \frac{N}{P (N - P)} d P = \int k d t$

attempts to write $\frac{N}{P (N - P)}$ in partial fractions form M1

$\frac{N}{P (N - P)} \equiv \frac{A}{P} + \frac{B}{(N - P)} \Rightarrow N \equiv A (N - P) + B P$

$A = 1, B = 1$ A1

$\frac{N}{P (N - P)} \equiv \frac{1}{P} + \frac{1}{(N - P)}$

$\int (\frac{1}{P} + \frac{1}{(N - P)}) d P = \int k d t$

$\Rightarrow \ln P - \ln (N - P) = k t (+ C)$ A1A1

Note: Award A1 for $- \ln (N - P)$ and A1 for $\ln P$ and $k t (+ C)$ . Absolute value signs are not required.

attempts to find $C$ in terms of $N$ and $P_{0}$ M1

when $t = 0, P = P_{0}$ and so $C = \ln P_{0} - \ln (N - P_{0})$

$k t = \ln (\frac{P}{N - P}) - \ln (\frac{P_{0}}{N - P_{o}}) (= \ln (\frac{\frac{P}{N - P}}{\frac{P_{0}}{N - P_{0}}}))$ A1

so $k t = \ln \frac{P}{P_{0}} (\frac{N - P_{0}}{N - P})$ AG

METHOD 2

attempts to separate variables M1

$\int \frac{1}{P (1 - \frac{P}{N})} d P = \int k d t$

attempts to write $\frac{1}{P (1 - \frac{P}{N})}$ in partial fractions form M1

$\frac{1}{P (1 - \frac{P}{N})} \equiv \frac{A}{P} + \frac{B}{1 - \frac{P}{N}} \Rightarrow 1 \equiv A (1 - \frac{P}{N}) + B P$

$A = 1, B = \frac{1}{N}$ A1

$\frac{1}{P (1 - \frac{P}{N})} \equiv \frac{1}{P} + \frac{1}{N (1 - \frac{P}{N})}$

$\int \frac{1}{P} + \frac{1}{N (1 - \frac{P}{N})} d P = \int k d t$

$\Rightarrow \ln P - \ln (1 - \frac{P}{N}) = k t (+ C)$ A1A1

Note: Award A1 for $- \ln (1 - \frac{P}{N})$ and A1 for $\ln P$ and $k t (+ C)$ . Absolute value signs are not required.

$\ln (\frac{P}{1 - \frac{P}{N}}) = k t + C \Rightarrow \ln (\frac{N P}{N - P}) = k t + C$

attempts to find $C$ in terms of $N$ and $P_{0}$ M1

when $t = 0, P = P_{0}$ and so $C = \ln (\frac{N P_{0}}{N - P_{0}})$

$k t = \ln (\frac{N P}{N - P}) - \ln (\frac{N P_{0}}{N - P_{0}}) (= \ln \frac{\frac{P}{N - P}}{\frac{P_{0}}{N - P_{0}}})$ A1

$k t = \ln \frac{P}{P_{0}} (\frac{N - P_{0}}{N - P})$ AG

METHOD 3

lets $u = \frac{1}{P}$ and forms $\frac{d u}{d t} = - \frac{1}{P^{2}} \frac{d P}{d t}$ M1

multiplies both sides of the differential equation by $- \frac{1}{P^{2}}$ and makes the above substitutions M1

$- \frac{1}{P^{2}} \frac{d P}{d t} = k (\frac{1}{N} - \frac{1}{P}) \Rightarrow \frac{d u}{d t} = k (\frac{1}{N} - u)$

$\frac{d u}{d t} + k u = \frac{k}{N}$ (linear first-order DE) A1

$IF = e^{\int k d t} = e^{k t} \Rightarrow e^{k t} \frac{d u}{d t} + k e^{k t} u = \frac{k}{N} e^{k t}$ (M1)

$\frac{d}{d t} (u e^{k t}) = \frac{k}{N} e^{k t}$

$u e^{k t} = \frac{1}{N} e^{k t} (+ C) (\frac{1}{P} e^{k t} = \frac{1}{N} e^{k t} (+ C))$ A1

attempts to find $C$ in terms of $N$ and $P_{0}$ M1

when $t = 0, P = P_{0}, u = \frac{1}{P_{0}}$ and so $C = \frac{1}{P_{0}} - \frac{1}{N} (= \frac{N - P_{0}}{N P_{0}})$

$e^{k t} (\frac{N - P}{N P}) = \frac{N - P_{0}}{N P_{0}}$

$e^{k t} = (\frac{P}{N - P}) (\frac{N - P_{0}}{P_{0}})$ A1

$k t = \ln \frac{P}{P_{0}} (\frac{N - P_{0}}{N - P})$ AG

[7 marks]

substitutes $t = 10, P = 3 P_{0}$ and $N = 4 P_{0}$ into $k t = \ln \frac{P}{P_{0}} (\frac{N - P_{0}}{N - P})$ M1

$10 k = \ln 3 (\frac{4 P_{0} - P_{0}}{4 P_{0} - 3 P_{0}}) (= \ln 9)$

$k = 0.220 (= \frac{1}{10} \ln 9, = \frac{1}{5} \ln 3)$ A1

[2 marks]

Examiners report

An extremely tricky question even for the strong candidates. Many struggled to understand what was expected in parts (b) and (c). As the question was set all with pronumerals instead of numbers many candidates found it challenging, thrown at deep water for parts (b), (c) and (e). It definitely was the question to show their skills for the Level 7 candidates provided that they did not run out of time.

Part (a) Very well answered, mostly correctly referring to the rate of change. Some candidates did not gain this mark because their sentence did not include the reference to the rate of change. Worded explanations continue being problematic to many candidates.

Part (b) Many candidates were confused how to approach this question and did not realise that they
needed to differentiate implicitly. Some tried but with errors, some did not fully show what was required.

Part (c) Most candidates started with equating the second derivative to zero. Most gave the answer $P = \frac{N}{2}$ omitting the other two possibilities. Most stopped here. Only a small number of candidates provided the correct mathematical argument to show it is a local maximum.

Part (d) Well done by those candidates who got that far. Most got the correct answer, sometimes not fully simplified.

Part (e) Most candidates separated the variables, but some were not able to do much more. Some candidates knew to resolve into partial fractions and attempted to do so, mainly successfully. Then they integrated, again, mainly successfully and continued to substitute the initial condition and manipulate the equation accordingly.

Part (f) Algebraic manipulation of the logarithmic expression was too much for some candidates with a common error of 0.33 given as the answer. The strong candidates provided the correct exact or rounded value.

[N/A]

EXN.2.AHL.TZ0.6a

Question

The curve $C$ has equation $e^{2 y} = x^{3} + y$ .

Show that $\frac{d y}{d x} = \frac{3 x^{2}}{2 e^{2 y} - 1}$ .

[3]

The tangent to $C$ at the point Ρ is parallel to the $y$ -axis.

Find the $x$ -coordinate of Ρ.

[4]

Markscheme

attempts implicit differentiation on both sides of the equation M1

$2 e^{2 y} \frac{d y}{d x} = 3 x^{2} + \frac{d y}{d x}$ A1

$(2 e^{2 y} - 1) \frac{d y}{d x} = 3 x^{2}$ A1

so $\frac{d y}{d x} = \frac{3 x^{2}}{2 e^{2 y} - 1}$ AG

[3 marks]

attempts to solve $2 e^{2 y} - 1 = 0$ for $y$ (M1)

$y = - 0.346 \dots (= \frac{1}{2} \ln \frac{1}{2})$ A1

attempts to solve $e^{2 y} = x^{3} + y$ for $x$ given their value of $y$ (M1)

$x = 0.946 (= {(\frac{1}{2} (1 - \ln \frac{1}{2}))}^{\frac{1}{3}})$ A1

[4 marks]

Examiners report

[N/A]

EXN.2.AHL.TZ0.12a

Question

Consider the differential equation

$\frac{d y}{d x} = f (\frac{y}{x}), x > 0$

The curve $y = f (x)$ for $x > 0$ has a gradient function given by

$\frac{d y}{d x} = \frac{y^{2} + 3 x y + 2 x^{2}}{x^{2}}$ .

The curve passes through the point $(1, - 1)$ .

Use the substitution $y = v x$ to show that $\int \frac{d v}{f (v) - v} = \ln x + C$ where $C$ is an arbitrary constant.

[3]

By using the result from part (a) or otherwise, solve the differential equation and hence show that the curve has equation $y = x (\tan (\ln x) - 1)$ .

[9]

The curve has a point of inflexion at $(x_{1}, y_{1})$ where $e^{- \frac{π}{2}} < x_{1} < e^{\frac{π}{2}}$ . Determine the coordinates of this point of inflexion.

[6]

Use the differential equation $\frac{d y}{d x} = \frac{y^{2} + 3 x y + 2 x^{2}}{x^{2}}$ to show that the points of zero gradient on the curve lie on two straight lines of the form $y = m x$ where the values of $m$ are to be determined.

[4]

Markscheme

$y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$ M1

$v + x \frac{d v}{d x} = f (v)$ A1

$\int \frac{d v}{f (v) - v} = \int \frac{d x}{x}$ A1

integrating the RHS, $\int \frac{d v}{f (v) - v} = \ln x + C$ AG

[3 marks]

EITHER

attempts to find $f (v)$ M1

$f (v) = v^{2} + 3 v + 2$ (A1)

substitutes their $f (v)$ into $\int \frac{d v}{f (v) - v}$ M1

$\int \frac{d v}{f (v) - v} = \int \frac{d v}{v^{2} + 2 v + 2}$

attempts to complete the square (M1)

$\int \frac{d v}{{(v + 1)}^{2} + 1}$ A1

$arctan (v + 1) (= \ln x + C)$ A1

attempts to find $f (v)$ M1

$v + x \frac{d v}{d x} = v^{2} + 3 v + 2$ A1

$\int \frac{d v}{v^{2} + 2 v + 2} = \frac{d x}{x}$ M1

attempts to complete the square (M1)

$\int \frac{d v}{{(v + 1)}^{2} + 1} (= \int \frac{d x}{x})$ A1

$arctan (v + 1) (= \ln x + C)$ A1

THEN

when $x = 1$ , $v = - 1$ (or $y = - 1$ ) and so $C = 0$ M1

substitutes for $v$ into their expression M1

$arctan (\frac{y}{x} + 1) = \ln x$

$\frac{y}{x} + 1 = \tan (\ln x)$ A1

so $y = x (\tan (\ln x) - 1)$ AG

[9 marks]

METHOD 1

EITHER

a correct graph of $y = f' (x)$ (for approximately $e^{- \frac{π}{2}} < x < e^{\frac{π}{2}}$ ) with a local minimum point below the $x$ -axis A2

Note: Award M1A1 for $\frac{d y}{d x} = \tan (\ln x) + {sec}^{2} (\ln x) - 1$ .

attempts to find the $x$ -coordinate of the local minimum point on the graph of $y = f' (x)$ (M1)

a correct graph of $y = f'' (x)$ (for approximately $e^{- \frac{π}{2}} < x < e^{\frac{π}{2}}$ ) showing the location of the $x$ -intercept A2

Note: Award M1A1 for $\frac{d^{2} y}{d x^{2}} = \frac{{sec}^{2} (\ln x)}{x} + \frac{2 {sec}^{2} (\ln x) \tan (\ln x)}{x}$ .

attempts to find the $x$ -intercept (M1)

THEN

$x = 0.629 (= e^{- arctan \frac{1}{2}})$ A1

attempts to find $f (0.629) (f (e^{- arctan \frac{1}{2}}))$ (M1)

the coordinates are $(0.629, - 0.943) (e^{- arctan \frac{1}{2}}, - \frac{3}{2} e^{- arctan \frac{1}{2}})$ A1

METHOD 2

attempts implicit differentiation on $\frac{d y}{d x}$ to find $\frac{d^{2} y}{d x^{2}}$ M1

$\frac{d^{2} y}{d x^{2}} = \frac{(2 y + 3 x) (x \frac{d y}{d x} - y)}{x^{3}}$ (or equivalent)

$\frac{d^{2} y}{d x^{2}} = 0 \Rightarrow y = - \frac{3 x}{2}$ ( $\frac{d y}{d x} \neq \frac{y}{x}$ ) A1

attempts to solve $- \frac{3 x}{2} = x (\tan (\ln x) - 1)$ for $x$ where $e^{- \frac{π}{2}} < x < e^{\frac{π}{2}}$ M1

$x = 0.629 (= e^{- arctan \frac{1}{2}})$ A1

attempts to find $f (0.629) (f (= e^{- arctan \frac{1}{2}}))$ (M1)

the coordinates are $(0.629, - 0.943) (e^{- arctan \frac{1}{2}}, - \frac{3}{2} e^{- arctan \frac{1}{2}})$ A1

[6 marks]

$\frac{d y}{d x} = 0 \Rightarrow y^{2} + 3 x y + 2 x^{2} = 0$ M1

attempts to solve $y^{2} + 3 x y + 2 x^{2} = 0$ for $y$ M1

$(y + 2 x) (y + x) = 0$ or $y = \frac{- 3 x \pm \sqrt{{(3 x)}^{2} - 4 (2 x^{2})}}{2} (= \frac{- 3 x \pm x}{2}, (x > 0))$ A1

$y = - 2 x$ and $y = - x (m = - 2, - 1)$ A1

Note: Award M1 for stating $\frac{d y}{d x} = 0$ , M1 for substituting $y = m x$ into $\frac{d y}{d x} (= 0)$ , A1 for $(m + 2) (m + 1) = 0$ and A1 for $m = - 2, - 1 \Rightarrow y = - 2 x$ and $y = - x$ .

[4 marks]

Examiners report

[N/A]

SPM.2.AHL.TZ0.11a

Question

A large tank initially contains pure water. Water containing salt begins to flow into the tank The solution is kept uniform by stirring and leaves the tank through an outlet at its base. Let $x$ grams represent the amount of salt in the tank and let $t$ minutes represent the time since the salt water began flowing into the tank.

The rate of change of the amount of salt in the tank, $\frac{d x}{d t}$ , is described by the differential equation $\frac{d x}{d t} = 10 e^{- \frac{t}{4}} - \frac{x}{t + 1}$ .

Show that $t$ + 1 is an integrating factor for this differential equation.

[2]

Hence, by solving this differential equation, show that $x (t) = \frac{200 - 40 e^{- \frac{t}{4}} (t + 5)}{t + 1}$ .

[8]

Sketch the graph of $x$ versus $t$ for 0 ≤ $t$ ≤ 60 and hence find the maximum amount of salt in the tank and the value of $t$ at which this occurs.

[5]

Find the value of $t$ at which the amount of salt in the tank is decreasing most rapidly.

[2]

The rate of change of the amount of salt leaving the tank is equal to $\frac{x}{t + 1}$ .

Find the amount of salt that left the tank during the first 60 minutes.

[4]

Markscheme

METHOD 1

$I (t) = e^{\int P (t) d t}$ M1

$e^{\int \frac{1}{t + 1} d t}$

= $e^{ln (t + 1)}$ A1

$= t + 1$ AG

METHOD 2

attempting product rule differentiation on $\frac{d}{d t} (x (t + 1))$ M1

$\frac{d}{d t} (x (t + 1)) = \frac{d x}{d t} (t + 1) + x$

$= (t + 1) (\frac{d x}{d t} + \frac{x}{t + 1})$ A1

so $t + 1$ is an integrating factor for this differential equation AG

[2 marks]

attempting to multiply through by $(t + 1)$ and rearrange to give (M1)

$(t + 1) \frac{d x}{d t} + x = 10 (t + 1) e^{- \frac{t}{4}}$ A1

$\frac{d}{d t} (x (t + 1)) = 10 (t + 1) e^{- \frac{t}{4}}$

$x (t + 1) = \int 10 (t + 1) e^{- \frac{t}{4}} d t$ A1

attempting to integrate the RHS by parts M1

$= - 40 (t + 1) e^{- \frac{t}{4}} + 40 \int e^{- \frac{t}{4}} d t$

$= - 40 (t + 1) e^{- \frac{t}{4}} - 160 e^{- \frac{t}{4}} + C$ A1

Note: Condone the absence of C.

EITHER

substituting $t = 0, x = 0 \Rightarrow C = 200$ M1

$x = \frac{- 40 (t + 1) e^{- \frac{t}{4}} - 160 e^{- \frac{t}{4}} + 200}{t + 1}$ A1

using $- 40 e^{- \frac{t}{4}}$ as the highest common factor of $- 40 (t + 1) e^{- \frac{t}{4}}$ and $- 160 e^{- \frac{t}{4}}$ M1

using $- 40 e^{- \frac{t}{4}}$ as the highest common factor of $- 40 (t + 1) e^{- \frac{t}{4}}$ and $- 160 e^{- \frac{t}{4}}$ giving

$x (t + 1) = - 40 e^{- \frac{t}{4}} (t + 5) + C$ (or equivalent) M1A1

substituting $t = 0, x = 0 \Rightarrow C = 200$ M1

THEN

$x (t) = \frac{200 - 40 e^{- \frac{t}{4}} (t + 5)}{t + 1}$ AG

[8 marks]

graph starts at the origin and has a local maximum (coordinates not required) A1

sketched for 0 ≤ $t$ ≤ 60 A1

correct concavity for 0 ≤ $t$ ≤ 60 A1

maximum amount of salt is 14.6 (grams) at $t$ = 6.60 (minutes) A1A1

[5 marks]

using an appropriate graph or equation (first or second derivative) M1

amount of salt is decreasing most rapidly at $t$ = 12.9 (minutes) A1

[2 marks]

EITHER

attempting to form an integral representing the amount of salt that left the tank M1

$\int_{0}^{60} \frac{x (t)}{t + 1} d t$

$\int_{0}^{60} \frac{200 - 40 e^{- \frac{t}{4}} (t + 5)}{{(t + 1)}^{2}} d t$ A1

attempting to form an integral representing the amount of salt that entered the tank minus the amount of salt in the tank at $t$ = 60(minutes)

amount of salt that left the tank is $\int_{0}^{60} 10 e^{- \frac{t}{4}} d t - x (60)$ A1

THEN

= 36.7 (grams) A2

[4 marks]

Examiners report

[N/A]

16N.3.AHL.TZ0.Hca_1a

Question

Consider the differential equation $\frac{d y}{d x} + (\frac{2 x}{1 + x^{2}}) y = x^{2}$ , given that $y = 2$ when $x = 0$ .

Show that $1 + x^{2}$ is an integrating factor for this differential equation.

[5]

Hence solve this differential equation. Give the answer in the form $y = f (x)$ .

[6]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

attempting to find an integrating factor (M1)

$\int \frac{2 x}{1 + x^{2}} d x = \ln (1 + x^{2})$ (M1)A1

IF is $e^{\ln (1 + x^{2})}$ (M1)A1

$= 1 + x^{2}$ AG

METHOD 2

multiply by the integrating factor

$(1 + x^{2}) \frac{d y}{d x} + 2 x y = x^{2} (1 + x^{2})$ M1A1

left hand side is equal to the derivative of $(1 + x^{2}) y$

[5 marks]

$(1 + x^{2}) \frac{d y}{d x} + 2 x y = (1 + x^{2}) x^{2}$ (M1)

$\frac{d}{d x} [(1 + x^{2}) y] = x^{2} + x^{4}$

$(1 + x^{2}) y = (\int x^{2} + x^{4} d x =) \frac{x^{3}}{3} + \frac{x^{5}}{5} (+ c)$ A1A1

$y = \frac{1}{1 + x^{2}} (\frac{x^{3}}{3} + \frac{x^{5}}{5} + c)$

$x = 0, y = 2 \Rightarrow c = 2$ M1A1

$y = \frac{1}{1 + x^{2}} (\frac{x^{3}}{3} + \frac{x^{5}}{5} + 2)$ A1

[6 marks]

Examiners report

[N/A]

17M.3.AHL.TZ0.Hca_1

Question

Use l’Hôpital’s rule to determine the value of

$lim_{x \to 0} \frac{\sin^{2} x}{x \ln (1 + x)} .$

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to use l’Hôpital’s rule, M1

$limit = lim_{x \to 0} \frac{2 \sin x \cos x}{\ln (1 + x) + \frac{x}{1 + x}}$ or $\frac{\sin 2 x}{\ln (1 + x) + \frac{x}{1 + x}}$ A1A1

Note: Award A1 for numerator A1 for denominator.

this gives 0/0 so use the rule again (M1)

$= lim_{x \to 0} \frac{2 \cos^{2} x - 2 \sin^{2} x}{\frac{1}{1 + x} + \frac{1 + x - x}{{(1 + x)}^{2}}}$ or $\frac{2 \cos 2 x}{\frac{2 + x}{{(1 + x)}^{2}}}$ A1A1

Note: Award A1 for numerator A1 for denominator.

$= 1$ A1

Note: This A1 is dependent on all previous marks being awarded, except when the first application of L’Hopital’s does not lead to 0/0, when it should be awarded for the correct limit of their derived function.

[7 marks]

Examiners report

[N/A]

17M.3.AHL.TZ0.Hca_4a

Question

Consider the differential equation

$\frac{d y}{d x} = f (\frac{y}{x}), x > 0.$

Use the substitution $y = v x$ to show that the general solution of this differential equation is

$\int \frac{d v}{f (v) - v} = \ln x + Constant.$

[3]

Hence, or otherwise, solve the differential equation

$\frac{d y}{d x} = \frac{x^{2} + 3 x y + y^{2}}{x^{2}}, x > 0,$

given that $y = 1$ when $x = 1$ . Give your answer in the form $y = g (x)$ .

[10]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$ M1

the differential equation becomes

$v + x \frac{d v}{d x} = f (v)$ A1

$\int \frac{d v}{f (v) - v} = \int \frac{d v}{x}$ A1

integrating, Constant $\int \frac{d v}{f (v) - v} = \ln x + Constant$ AG

[3 marks]

EITHER

$f (v) = 1 + 3 v + v^{2}$ (A1)

$(\int \frac{d v}{f (v) - v} =) \int \frac{d v}{1 + 3 v + v^{2} - v} = \ln x + C$ M1A1

$\int \frac{d v}{{(1 + v)}^{2}} = (\ln x + C)$ A1

Note: A1 is for correct factorization.

$- \frac{1}{1 + v} (= \ln x + C)$ A1

$v + x \frac{d v}{d x} = 1 + 3 v + v^{2}$ A1

$\int \frac{d v}{1 + 2 v + v^{2}} = \int \frac{1}{x} d x$ M1

$\int \frac{d v}{{(1 + v)}^{2}} (= \int \frac{1}{x} d x)$ (A1)

Note: A1 is for correct factorization.

$- \frac{1}{1 + v} = \ln x (+ C)$ A1A1

THEN

substitute $y = 1$ or $v = 1$ when $x = 1$ (M1)

therefore $C = - \frac{1}{2}$ A1

Note: This A1 can be awarded anywhere in their solution.

substituting for $v$ ,

$- \frac{1}{(1 + \frac{y}{x})} = \ln x - \frac{1}{2}$ M1

Note: Award for correct substitution of $\frac{y}{x}$ into their expression.

$1 + \frac{y}{x} = \frac{1}{\frac{1}{2} - \ln x}$ (A1)

Note: Award for any rearrangement of a correct expression that has $y$ in the numerator.

$y = x (\frac{1}{(\frac{1}{2} - \ln x)} - 1) (orequivalent)$ A1

$(= x (\frac{1 + 2 \ln x}{1 - 2 \ln x}))$

[10 marks]

Examiners report

[N/A]

17N.3.AHL.TZ0.Hca_2a

Question

Consider the differential equation $\frac{d y}{d x} + \frac{x}{x^{2} + 1} y = x$ where $y = 1$ when $x = 0$ .

Show that $\sqrt{x^{2} + 1}$ is an integrating factor for this differential equation.

[4]

Solve the differential equation giving your answer in the form $y = f (x)$ .

[6]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

integrating factor $= e^{\int \frac{x}{x^{2} + 1} d x}$ (M1)

$\int \frac{x}{x^{2} + 1} d x = \frac{1}{2} \ln (x^{2} + 1)$ (M1)

Note: Award M1 for use of $u = x^{2} + 1$ for example or $\int \frac{f^{'} (x)}{f (x)} d x = \ln f (x)$ .

integrating factor $= e^{\frac{1}{2} \ln (x^{2} + 1)}$ A1

$= e^{\ln (\sqrt{x^{2} + 1})}$ A1

Note: Award A1 for $e^{\ln \sqrt{u}}$ where $u = x^{2} + 1$ .

$= \sqrt{x^{2} + 1}$ AG

METHOD 2

$\frac{d}{d x} (y \sqrt{x^{2} + 1}) = \frac{d y}{d x} \sqrt{x^{2} + 1} + \frac{x}{\sqrt{x^{2} + 1}} y$ M1A1

$\sqrt{x^{2} + 1} (\frac{d y}{d x} + \frac{x}{x^{2} + 1} y)$ M1A1

Note: Award M1 for attempting to express in the form $\sqrt{x^{2} + 1} \times (LHSofde)$ .

so $\sqrt{x^{2} + 1}$ is an integrating factor for this differential equation AG

[4 marks]

$\sqrt{x^{2} + 1} \frac{d y}{d x} + \frac{x}{\sqrt{x^{2} + 1}} y = x \sqrt{x^{2} + 1}$ (or equivalent) (M1)

$\frac{d}{d x} (y \sqrt{x^{2} + 1}) = x \sqrt{x^{2} + 1}$

$y \sqrt{x^{2} + 1} = \int x \sqrt{x^{2} + 1} d x (y = \frac{1}{\sqrt{x^{2} + 1}} \int x \sqrt{x^{2} + 1} d x)$ A1

$= \frac{1}{3} (x^{2} + 1)^{\frac{3}{2}} + C$ (M1)A1

Note: Award M1 for using an appropriate substitution.

Note: Condone the absence of $C$ .

substituting $x = 0, y = 1 \Rightarrow C = \frac{2}{3}$ M1

Note: Award M1 for attempting to find their value of $C$ .

$y = \frac{1}{3} (x^{2} + 1) + \frac{2}{3 \sqrt{x^{2} + 1}} (y = \frac{{(x^{2} + 1)}^{\frac{3}{2}} + 2}{3 \sqrt{x^{2} + 1}})$ A1

[6 marks]

Examiners report

[N/A]

18M.3.AHL.TZ0.Hca_3a

Question

Find the value of $\int_{4}^{\infty} \frac{1}{x^{3}} d x$ .

[3]

Illustrate graphically the inequality $\sum_{n = 5}^{\infty} \frac{1}{n^{3}} < \int_{4}^{\infty} \frac{1}{x^{3}} d x < \sum_{n = 4}^{\infty} \frac{1}{n^{3}}$ .

[4]

Hence write down a lower bound for $\sum_{n = 4}^{\infty} \frac{1}{n^{3}}$ .

[1]

Find an upper bound for $\sum_{n = 4}^{\infty} \frac{1}{n^{3}}$ .

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\int_{4}^{\infty} \frac{1}{x^{3}} d x = lim_{R \to \infty} \int_{4}^{R} \frac{1}{x^{3}} d x$ (A1)

Note: The above A1 for using a limit can be awarded at any stage.
Condone the use of $lim_{x \to \infty}$ .

Do not award this mark to candidates who use $\infty$ as the upper limit throughout.

= $lim_{R \to \infty} {[- \frac{1}{2} x^{- 2}]}_{4}^{R} (= {[- \frac{1}{2} x^{- 2}]}_{4}^{\infty})$ M1

$= lim_{R \to \infty} (- \frac{1}{2} (R^{- 2} - 4^{- 2}))$

$= \frac{1}{32}$ A1

[3 marks]

A1A1A1A1

A1 for the curve
A1 for rectangles starting at $x = 4$
A1 for at least three upper rectangles
A1 for at least three lower rectangles

Note: Award A0A1 for two upper rectangles and two lower rectangles.

sum of areas of the lower rectangles < the area under the curve < the sum of the areas of the upper rectangles so

$\sum_{n = 5}^{\infty} \frac{1}{n^{3}} < \int_{4}^{\infty} \frac{1}{x^{3}} d x < \sum_{n = 4}^{\infty} \frac{1}{n^{3}}$ AG

[4 marks]

a lower bound is $\frac{1}{32}$ A1

Note: Allow FT from part (a).

[1 mark]

METHOD 1

$\sum_{n = 5}^{\infty} \frac{1}{n^{3}} < \frac{1}{32}$ (M1)

$\frac{1}{64} + \sum_{n = 5}^{\infty} \frac{1}{n^{3}} = \frac{1}{32} + \frac{1}{64}$ (M1)

$\sum_{n = 4}^{\infty} \frac{1}{n^{3}} < \frac{3}{64}$ , an upper bound A1

Note: Allow FT from part (a).

METHOD 2

changing the lower limit in the inequality in part (b) gives

$\sum_{n = 4}^{\infty} \frac{1}{n^{3}} < \int_{3}^{\infty} \frac{1}{x^{3}} d x (< \sum_{n = 3}^{\infty} \frac{1}{n^{3}})$ (A1)

$\sum_{n = 4}^{\infty} \frac{1}{n^{3}} < lim_{R \to \infty} {[- \frac{1}{2} x^{- 2}]}_{3}^{R}$ (M1)

$\sum_{n = 4}^{\infty} \frac{1}{n^{3}} < \frac{1}{18}$ , an upper bound A1

Note: Condone candidates who do not use a limit.

[3 marks]

Examiners report

[N/A]

18M.3.AHL.TZ0.Hca_4a

Question

The function $f$ is defined by $f (x) = (\arcsin x)^{2}, - 1 ⩽ x ⩽ 1$ .

The function $f$ satisfies the equation $(1 - x^{2}) f^{″} (x) - x f^{'} (x) - 2 = 0$ .

Show that $f^{'} (0) = 0$ .

[2]

By differentiating the above equation twice, show that

$(1 - x^{2}) f^{(4)} (x) - 5 x f^{(3)} (x) - 4 f^{″} (x) = 0$

where $f^{(3)} (x)$ and $f^{(4)} (x)$ denote the 3rd and 4th derivative of $f (x)$ respectively.

[4]

Hence show that the Maclaurin series for $f (x)$ up to and including the term in $x^{4}$ is $x^{2} + \frac{1}{3} x^{4}$ .

[3]

Use this series approximation for $f (x)$ with $x = \frac{1}{2}$ to find an approximate value for $π^{2}$ .

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$f^{'} (x) = \frac{2 arcsin (x)}{\sqrt{1 - x^{2}}}$ M1A1

Note: Award M1 for an attempt at chain rule differentiation.
Award M0A0 for $f^{'} (x) = 2 arcsin (x)$ .

$f^{'} (0) = 0$ AG

[2 marks]

differentiating gives $(1 - x^{2}) f^{(3)} (x) - 2 x f^{″} (x) - f^{'} (x) - x f^{″} (x) (= 0)$ M1A1

differentiating again gives $(1 - x^{2}) f^{(4)} (x) - 2 x f^{(3)} (x) - 3 f^{″} (x) - 3 x f^{(3)} (x) - f^{″} (x) (= 0)$ M1A1

Note: Award M1 for an attempt at product rule differentiation of at least one product in each of the above two lines.
Do not penalise candidates who use poor notation.

$(1 - x^{2}) f^{(4)} (x) - 5 x f^{(3)} (x) - 4 f^{″} (x) = 0$ AG

[4 marks]

attempting to find one of $f^{″} (0)$ , $f^{(3)} (0)$ or $f^{(4)} (0)$ by substituting $x = 0$ into relevant differential equation(s) (M1)

Note: Condone $f^{″} (0)$ found by calculating $\frac{d}{d x} (\frac{2 arcsin (x)}{\sqrt{1 - x^{2}}})$ at $x = 0$ .

$(f (0) = 0, f^{'} (0) = 0)$

$f^{″} (0) = 2$ and $f^{(4)} (0) - 4 f^{″} (0) = 0 \Rightarrow f^{(4)} (0) = 8$ A1

$f^{(3)} (0) = 0$ and so $\frac{2}{2!} x^{2} + \frac{8}{4!} x^{4}$ A1

Note: Only award the above A1, for correct first differentiation in part (b) leading to $f^{(3)} (0) = 0$ stated or $f^{(3)} (0) = 0$ seen from use of the general Maclaurin series.
Special Case: Award (M1)A0A1 if $f^{(4)} (0) = 8$ is stated without justification or found by working backwards from the general Maclaurin series.

so the Maclaurin series for $f (x)$ up to and including the term in $x^{4}$ is $x^{2} + \frac{1}{3} x^{4}$ AG

[3 marks]

substituting $x = \frac{1}{2}$ into $x^{2} + \frac{1}{3} x^{4}$ M1

the series approximation gives a value of $\frac{13}{48}$

so $π^{2} ≃ \frac{13}{48} \times 36$

$≃ 9.75 (≃ \frac{39}{4})$ A1

Note: Accept 9.76.

[2 marks]

Examiners report

[N/A]

18M.3.AHL.TZ0.Hca_5a

Question

Consider the differential equation $x \frac{d y}{d x} - y = x^{p} + 1$ where $x \in R, x \neq 0$ and $p$ is a positive integer, $p > 1$ .

Solve the differential equation given that $y = - 1$ when $x = 1$ . Give your answer in the form $y = f (x)$ .

[8]

Show that the $x$ -coordinate(s) of the points on the curve $y = f (x)$ where $\frac{d y}{d x} = 0$ satisfy the equation $x^{p - 1} = \frac{1}{p}$ .

[2]

b.i.

Deduce the set of values for $p$ such that there are two points on the curve $y = f (x)$ where $\frac{d y}{d x} = 0$ . Give a reason for your answer.

[2]

b.ii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

$\frac{d y}{d x} = \frac{y}{x} = x^{p - 1} + \frac{1}{x}$ (M1)

integrating factor $= e^{\int - \frac{1}{x} d x}$ M1

$= e^{- ln x}$ (A1)

= $\frac{1}{x}$ A1

$\frac{1}{x} \frac{d y}{d x} - \frac{y}{x^{2}} = x^{p - 2} + \frac{1}{x^{2}}$ (M1)

$\frac{d}{d x} (\frac{y}{x}) = x^{p - 2} + \frac{1}{x^{2}}$

$\frac{y}{x} = \frac{1}{p - 1} x^{p - 1} - \frac{1}{x} + C$ A1

Note: Condone the absence of C.

$y = \frac{1}{p - 1} x^{p} + C x - 1$

substituting $x = 1$ , $y = - 1 \Rightarrow C = - \frac{1}{p - 1}$ M1

Note: Award M1 for attempting to find their value of C.

$y = \frac{1}{p - 1} (x^{p} - x) - 1$ A1

[8 marks]

METHOD 2

put $y = v x$ so that $\frac{d y}{d x} = v + x \frac{d v}{d x}$ M1(A1)

substituting, M1

$x (v + x \frac{d v}{d x}) - v x = x^{p} + 1$ (A1)

$x \frac{d v}{d x} = x^{p - 1} + \frac{1}{x}$ M1

$\frac{d v}{d x} = x^{p - 2} + \frac{1}{x^{2}}$

$v = \frac{1}{p - 1} x^{p - 1} - \frac{1}{x} + C$ A1

Note: Condone the absence of C.

$y = \frac{1}{p - 1} x^{p} + C x - 1$

substituting $x = 1$ , $y = - 1 \Rightarrow C = - \frac{1}{p - 1}$ M1

Note: Award M1 for attempting to find their value of C.

$y = \frac{1}{p - 1} (x^{p} - x) - 1$ A1

[8 marks]

METHOD 1

find $\frac{d y}{d x}$ and solve $\frac{d y}{d x} = 0$ for $x$

$\frac{d y}{d x} = \frac{1}{p - 1} (p x^{p - 1} - 1)$ M1

$\frac{d y}{d x} = 0 \Rightarrow p x^{p - 1} - 1 = 0$ A1

$p x^{p - 1} = 1$

Note: Award a maximum of M1A0 if a candidate’s answer to part (a) is incorrect.

$x^{p - 1} = \frac{1}{p}$ AG

METHOD 2

substitute $\frac{d y}{d x} = 0$ and their $y$ into the differential equation and solve for $x$

$\frac{d y}{d x} = 0 \Rightarrow - (\frac{x^{p} - x}{p - 1}) + 1 = x^{p} + 1$ M1

$x^{p} - x = x^{p} - p x^{p}$ A1

$p x^{p - 1} = 1$

Note: Award a maximum of M1A0 if a candidate’s answer to part (a) is incorrect.

$x^{p - 1} = \frac{1}{p}$ AG

[2 marks]

b.i.

there are two solutions for $x$ when $p$ is odd (and $p > 1$ A1

if $p - 1$ is even there are two solutions (to $x^{p - 1} = \frac{1}{p}$ )

and if $p - 1$ is odd there is only one solution (to $x^{p - 1} = \frac{1}{p}$ ) R1

Note: Only award the R1 if both cases are considered.

[4 marks]

b.ii.

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

18N.3.AHL.TZ0.Hca_2a

Question

Use L’Hôpital’s rule to determine the value of

$lim_{x \to 0} (\frac{{e^{- 3 x}}^{^{2}} + 3 cos (2 x) - 4}{3 x^{2}})$

[5]

Hence find $lim_{x \to 0} (\frac{\int_{0}^{x} ({e^{- 3 t}}^{^{2}} + 3 cos (2 t) - 4) d t}{\int_{0}^{x} 3 t^{2} d t})$ .

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$lim_{x \to 0} \frac{{e^{- 3 x}}^{^{2}} + 3 cos 2 x - 4}{3 x^{2}} = (\begin{matrix} 0 \\ 0 \end{matrix})$

$= lim_{x \to 0} \frac{- 6 x {e^{- 3 x}}^{^{2}} - 6 sin 2 x}{6 x} = (\begin{matrix} 0 \\ 0 \end{matrix})$ M1A1A1

$= lim_{x \to 0} \frac{- 6 {e^{- 3 x}}^{^{2}} + 36 x^{2} {e^{- 3 x}}^{^{2}} - 12 cos 2 x}{6}$ A1

= −3 A1

[5 marks]

$lim_{x \to 0} (\frac{\int_{0}^{x} ({e^{- 3 t}}^{^{2}} + 3 cos 2 t - 4) d t}{\int_{0}^{x} 3 t^{2} d t})$ is of the form $\frac{0}{0}$

applying l’Hôpital´s rule (M1)

$lim_{x \to 0} \frac{{e^{- 3 x}}^{^{2}} + 3 cos 2 x - 4}{3 x^{2}}$ (A1)

= −3 A1

[3 marks]

Examiners report

[N/A]

18N.3.AHL.TZ0.Hca_4a

Question

Consider the differential equation $\frac{d y}{d x} = 1 + \frac{y}{x}$ , where $x \neq 0$ .

Consider the family of curves which satisfy the differential equation $\frac{d y}{d x} = 1 + \frac{y}{x}$ , where $x \neq 0$ .

Given that $y (1) = 1$ , use Euler’s method with step length $h$ = 0.25 to find an approximation for $y (2)$ . Give your answer to two significant figures.

[4]

Solve the equation $\frac{d y}{d x} = 1 + \frac{y}{x}$ for $y (1) = 1$ .

[6]

Find the percentage error when $y (2)$ is approximated by the final rounded value found in part (a). Give your answer to two significant figures.

[3]

Find the equation of the isocline corresponding to $\frac{d y}{d x} = k$ , where $k \neq 0$ , $k \in R$ .

[1]

d.i.

Show that such an isocline can never be a normal to any of the family of curves that satisfy the differential equation.

[4]

d.ii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to apply Euler’s method (M1)

$x_{n + 1} = x_{n} + 0.25; y_{n + 1} = y_{n} + 0.25 \times (1 + \frac{y_{n}}{x_{n}})$

(A1)(A1)

Note: Award A1 for correct $x$ values, A1 for first three correct $y$ values.

$y$ = 3.3 A1

[4 marks]

METHOD 1

$I (x) = e^{\int - \frac{1}{x} d x}$ (M1)

$= e^{- ln x}$

$= \frac{1}{x}$ (A1)

$\frac{1}{x} \frac{d y}{d x} - \frac{y}{x^{2}} = \frac{1}{x}$ (M1)

$\frac{d}{d x} (\frac{y}{x}) = \frac{1}{x}$

$\frac{y}{x} = ln | x | + C$ A1

$y (1) = 1 \Rightarrow C = 1$ M1

$y = x ln | x | + x$ A1

METHOD 2

$v = \frac{y}{x}$ M1

$\frac{d v}{d x} = \frac{1}{x} \frac{d y}{d x} - \frac{1}{x^{2}} y$ (A1)

$v + x \frac{d v}{d x} = 1 + v$ M1

$\int 1 d v = \int \frac{1}{x} d x$

$v = ln | x | + C$

$\frac{y}{x} = ln | x | + C$ A1

$y (1) = 1 \Rightarrow C = 1$ M1

$y = x ln | x | + x$ A1

[6 marks]

$y (2) = 2 ln 2 + 2 = 3.38629 \dots$

percentage error $= \frac{3.38629 \dots - 3.3}{3.38629 \dots} \times 100 %$ (M1)(A1)

= 2.5% A1

[3 marks]

$\frac{d y}{d x} = k \Rightarrow 1 + \frac{y}{x} = k$ A1

$y = (k - 1) x$

[1 mark]

d.i.

gradient of isocline equals gradient of normal (M1)

$k - 1 = - \frac{1}{k}$ or $k (k - 1) = - 1$ A1

$k^{2} - k + 1 = 0$ A1

$Δ = 1 - 4 < 0$ R1

$∴$ no solution AG

Note: Accept alternative reasons for no solutions.

[4 marks]

d.ii.

Examiners report

[N/A]

d.i.

[N/A]

d.ii.

19M.3.AHL.TZ0.Hca_1a

Question

A simple model to predict the population of the world is set up as follows. At time $t$ years the population of the world is $x$ , which can be assumed to be a continuous variable. The rate of increase of $x$ due to births is 0.056 $x$ and the rate of decrease of $x$ due to deaths is 0.035 $x$ .

Show that $\frac{d x}{d t} = 0.021 x$ .

[1]

Find a prediction for the number of years it will take for the population of the world to double.

[6]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\frac{d x}{d t} = 0.056 x - 0.035 x$ A1

$\frac{d x}{d t} = 0.021 x$ AG

[1 mark]

METHOD 1

$\frac{d x}{d t} = 0.021 x$

attempt to separate variables M1

$\int \frac{1}{x} d x = \int 0.021 d t$ A1

$ln x = 0.021 t (+ c)$ A1

EITHER

$x = A e^{0.021 t}$

$\Rightarrow 2 A = A e^{0.021 t}$ A1

Note: This A1 is independent of the following marks.

$t = 0, x = x_{0} \Rightarrow c = ln x_{0}$

$\Rightarrow ln 2 x_{0} = 0.021 t + ln x_{0}$ A1

Note: This A1 is independent of the following marks.

THEN

$\Rightarrow ln 2 = 0.021 t$ (M1)

$\Rightarrow t = 33$ years A1

Note: If a candidate writes $t = 33.007$ , so $t = 34$ then award the final A1.

METHOD 2

$\frac{d x}{d t} = 0.021 x$

attempt to separate variables M1

$\int_{A}^{2 A} \frac{1}{x} d x = \int_{0}^{t} 0.021 d u$ A1A1

Note: Award A1 for correct integrals and A1 for correct limits seen anywhere. Do not penalize use of $t$ in place of $u$ .

${[ln x]}_{A}^{2 A} = {[0.021 u]}_{0}^{t}$ A1

$\Rightarrow ln 2 = 0.021 t$ (M1)

$\Rightarrow t = 33$ A1

[6 marks]

Examiners report

[N/A]

19M.3.AHL.TZ0.Hca_4

Question

Using L’Hôpital’s rule, find $lim_{x \to 0} (\frac{tan 3 x - 3 tan x}{sin 3 x - 3 sin x})$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$lim_{x \to 0} (\frac{tan 3 x - 3 tan x}{sin 3 x - 3 sin x})$

$lim_{x \to 0} (\frac{3 se c^{2} 3 x - 3 se c^{2} x}{3 cos 3 x - 3 cos x}) (= lim_{x \to 0} (\frac{se c^{2} 3 x - se c^{2} x}{cos 3 x - cos x}))$ M1A1A1

Note: Award M1 for attempt at differentiation using l'Hopital's rule, A1 for numerator, A1 for denominator.

METHOD 1

using l’Hopital’s rule again

$= lim_{x \to 0} (\frac{18 se c^{2} 3 x tan 3 x - 6 se c^{2} x tan x}{- 9 sin 3 x + 3 sin x}) (= lim_{x \to 0} (\frac{6 se c^{2} 3 x tan 3 x - 2 se c^{2} x tan x}{- 3 sin 3 x + sin x}))$ A1A1

EITHER

$= lim_{x \to 0} (\frac{108 se c^{2} 3 x ta n^{2} 3 x + 54 se c^{4} 3 x - 12 se c^{2} x ta n^{2} x - 6 se c^{4} x}{-27 cos 3 x + 3 cos x})$ A1A1

Note: Not all terms in numerator need to be written in final fraction. Award A1 for $54 se c^{4} 3 x + \dots - 6 se c^{4} x \dots -$ . However, if the terms are written, they
must be correct to award A1.

attempt to substitute $x = 0$ M1

$= \frac{48}{- 24}$

$\frac{d}{d x} (18 se c^{2} 3 x tan 3 x - 6 se c^{2} x tan x) |_{x = 0} = 48$ (M1)A1

$\frac{d}{d x} (- 9 sin 3 x + 3 sin x) |_{x = 0} = - 24$ A1

THEN

$(lim_{x \to 0} (\frac{tan 3 x - 3 tan x}{sin 3 x - 3 sin x})) = - 2$ A1

METHOD 2

$= lim_{x \to 0} (\frac{\frac{3}{co s^{2} 3 x} - \frac{3}{co s^{2} x}}{3 cos 3 x - 3 cos x})$ M1

$= lim_{x \to 0} (\frac{co s^{2} x - co s^{2} 3 x}{co s^{2} 3 x co s^{2} x (cos 3 x - cos x)})$ A1

$= lim_{x \to 0} (\frac{cos x + cos 3 x}{- co s^{2} 3 x co s^{2} x})$ M1A1

attempt to substitute $x = 0$ M1

$= \frac{2}{- 1}$

$= - 2$ A1

[9 marks]

Examiners report

[N/A]

19M.3.AHL.TZ0.Hca_5a

Question

Consider the differential equation $2 x y \frac{d y}{d x} = y^{2} - x^{2}$ , where $x > 0$ .

Solve the differential equation and show that a general solution is $x^{2} + y^{2} = c x$ where $c$ is a positive constant.

[11]

Prove that there are two horizontal tangents to the general solution curve and state their equations, in terms of $c$ .

[5]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\frac{d y}{d x} = \frac{y^{2} - x^{2}}{2 x y}$

let $y = v x$ M1

$\frac{d y}{d x} = v + x \frac{d v}{d x}$ (A1)

$v + x \frac{d v}{d x} = \frac{v^{2} x^{2} - x^{2}}{2 v x^{2}}$ (M1)

$v + x \frac{d v}{d x} = \frac{v^{2} - 1}{2 v} (= \frac{v}{2} - \frac{1}{2 v})$ (A1)

Note: Or equivalent attempt at simplification.

$x \frac{d v}{d x} = \frac{- v^{2} - 1}{2 v} (= - \frac{v}{2} - \frac{1}{2 v})$ A1

$\frac{2 v}{1 + v^{2}} \frac{d v}{d x} = - \frac{1}{x}$ (M1)

$\int \frac{2 v}{1 + v^{2}} d v = \int - \frac{1}{x} d x$ (A1)

$ln (1 + v^{2}) = - ln x + ln c$ A1A1

Note: Award A1 for LHS and A1 for RHS and a constant.

$ln (1 + {(\frac{y}{x})}^{2}) = - ln x + ln c$ M1

Note: Award M1 for substituting $v = \frac{y}{x}$ . May be seen at a later stage.

$1 + {(\frac{y}{x})}^{2} = \frac{c}{x}$ A1

Note: Award A1 for any correct equivalent equation without logarithms.

$x^{2} + y^{2} = c x$ AG

[11 marks]

METHOD 1

$\frac{d y}{d x} = \frac{y^{2} - x^{2}}{2 x y}$

(for horizontal tangents) $\frac{d y}{d x} = 0$ M1

$(\Rightarrow y^{2} = x^{2}) \Rightarrow y = \pm x$

EITHER

using $x^{2} + y^{2} = c x \Rightarrow 2 x^{2} = c x$ M1

$2 x^{2} - c x = 0 \Rightarrow x = \frac{c}{2}$ A1

Note: Award M1A1 for $2 y^{2} = \pm c y$ .

using implicit differentiation of $x^{2} + y^{2} = c x$

$2 x + 2 y \frac{d y}{d x} = c$ M1

Note: Accept differentiation of $y = \sqrt{c x - x^{2}}$ .

$\frac{d y}{d x} = 0 \Rightarrow x = \frac{c}{2}$ A1

THEN

tangents at $y = \frac{c}{2}, y = - \frac{c}{2}$ A1A1

hence there are two tangents AG

METHOD 2

$x^{2} + y^{2} = c x$

${(x - \frac{c}{2})}^{2} + y^{2} = \frac{c^{2}}{4}$ M1A1

this is a circle radius $\frac{c}{2}$ centre $(\frac{c}{2}, 0)$ A1

hence there are two tangents AG

tangents at $y = \frac{c}{2}, y = - \frac{c}{2}$ A1A1

[5 marks]

Examiners report

[N/A]

19N.3.AHL.TZ0.Hca_3a

Question

The function $f$ is defined by $f (x) = arcsin (2 x)$ , where $- \frac{1}{2} ⩽ x ⩽ \frac{1}{2}$ .

By finding a suitable number of derivatives of $f$ , find the first two non-zero terms in the Maclaurin series for $f$ .

[8]

Hence or otherwise, find $lim_{x \to 0} \frac{arcsin (2 x) - 2 x}{{(2 x)}^{3}}$ .

[3]

Markscheme

$f (x) = arcsin (2 x)$

$f^{'} (x) = \frac{2}{\sqrt{1 - 4 x^{2}}}$ M1A1

Note: Award M1A0 for $f^{'} (x) = \frac{1}{\sqrt{1 - 4 x^{2}}}$

$f^{″} (x) = \frac{8 x}{{(1 - 4 x^{2})}^{\frac{3}{2}}}$ A1

EITHER

$f^{‴} (x) = \frac{8 {(1 - 4 x^{2})}^{\frac{3}{2}} - 8 x (\frac{3}{2} (- 8 x) {(1 - 4 x^{2})}^{\frac{1}{2}})}{{(1 - 4 x^{2})}^{3}} (= \frac{8 {(1 - 4 x^{2})}^{\frac{3}{2}} + 96 x^{2} {(1 - 4 x^{2})}^{\frac{1}{2}}}{{(1 - 4 x^{2})}^{3}})$ A1

$f^{‴} (x) = 8 {(1 - 4 x^{2})}^{- \frac{3}{2}} + 8 x (- \frac{3}{2} {(1 - 4 x^{2})}^{- \frac{5}{2}}) (- 8 x) (= 8 {(1 - 4 x^{2})}^{- \frac{3}{2}} + 96 x^{2} {(1 - 4 x^{2})}^{- \frac{5}{2}})$ A1

THEN

substitute $x = 0$ into $f$ or any of its derivatives (M1)

$f (0) = 0$ , $f^{'} (0) = 2$ and $f^{″} (0) = 0$ A1

$f^{‴} (0) = 8$

the Maclaurin series is

$f (x) = 2 x + \frac{8 x^{3}}{6} + \dots (= 2 x + \frac{4 x^{3}}{3} + \dots)$ (M1)A1

[8 marks]

METHOD 1

$lim_{x \to 0} \frac{arcsin (2 x) - 2 x}{{(2 x)}^{3}} = lim_{x \to 0} \frac{2 x + \frac{4 x^{3}}{3} + \dots - 2 x}{8 x^{3}}$ M1

$= lim_{x \to 0} \frac{\frac{4}{3} + \dots termswith x}{8}$ (M1)

$= \frac{1}{6}$ A1

Note: Condone the omission of +… in their working.

METHOD 2

$lim_{x \to 0} \frac{arcsin (2 x) - 2 x}{{(2 x)}^{3}} = \frac{0}{0}$ indeterminate form, using L’Hôpital’s rule

$= lim_{x \to 0} \frac{\frac{2}{\sqrt{1 - 4 x^{2}}} - 2}{24 x^{2}}$ M1

$= \frac{0}{0}$ indeterminate form, using L’Hôpital’s rule again

$= lim_{x \to 0} \frac{\frac{8 x}{{(1 - 4 x^{2})}^{\frac{3}{2}}}}{48 x} (= lim_{x \to 0} \frac{1}{6 {(1 - 4 x^{2})}^{\frac{3}{2}}})$ M1

Note: Award M1 only if their previous expression is in indeterminate form.

$= \frac{1}{6}$ A1

Note: Award FT for use of their derivatives from part (a).

[3 marks]

Examiners report

[N/A]

19N.3.AHL.TZ0.Hca_4a

Question

Consider the differential equation $\frac{d y}{d x} = \frac{4 x^{2} + y^{2} - x y}{x^{2}}$ , with $y = 2$ when $x = 1$ .

Use Euler’s method, with step length $h = 0.1$ , to find an approximate value of $y$ when $x = 1.4$ .

[5]

Sketch the isoclines for $\frac{d y}{d x} = 4$ .

[3]

Express $m^{2} - 2 m + 4$ in the form ${(m - a)}^{2} + b$ , where $a, b \in Z$ .

[1]

c.i.

Solve the differential equation, for $x > 0$ , giving your answer in the form $y = f (x)$ .

[10]

c.ii.

Sketch the graph of $y = f (x)$ for $1 ⩽ x ⩽ 1.4$ .

[1]

c.iii.

With reference to the curvature of your sketch in part (c)(iii), and without further calculation, explain whether you conjecture $f (1.4)$ will be less than, equal to, or greater than your answer in part (a).

[2]

c.iv.

Markscheme

(M1)(A1)(A1)(A1)A1

$y (1.4) \approx 5.34$

Note: Award A1 for each correct $y$ value.
For the intermediate $y$ values, accept answers that are accurate to 2 significant figures.
The final $y$ value must be accurate to 3 significant figures or better.

[5 marks]

attempt to solve $\frac{4 x^{2} + y^{2} - x y}{x^{2}} = 4$ (M1)

$\Rightarrow y^{2} - x y = 0$

$y (y - x) = 0$

$y = 0$ or $y = x$

A1A1

[3 marks]

$m^{2} - 2 m + 4 = {(m - 1)}^{2} + 3 (a = 1, b = 3)$ A1

[1 mark]

c.i.

recognition of homogeneous equation,
let $y = v x$ M1

the equation can be written as

$v + x \frac{d v}{d x} = 4 + v^{2} - v$ (A1)

$x \frac{d v}{d x} = v^{2} - 2 v + 4$

$\int \frac{1}{v^{2} - 2 v + 4} d v = \int \frac{1}{x} d x$ M1

Note: Award M1 for attempt to separate the variables.

$\int \frac{1}{{(v - 1)}^{2} + 3} d v = \int \frac{1}{x} d x$ from part (c)(i) M1

$\frac{1}{\sqrt{3}} arctan (\frac{v - 1}{\sqrt{3}}) = ln x (+ c)$ A1A1

$x = 1, y = 2 \Rightarrow v = 2$

$\frac{1}{\sqrt{3}} arctan (\frac{1}{\sqrt{3}}) = ln 1 + c$ M1

Note: Award M1 for using initial conditions to find $c$ .

$\Rightarrow c = \frac{π}{6 \sqrt{3}} (= 0.302)$ A1

$arctan (\frac{v - 1}{\sqrt{3}}) = \sqrt{3} ln x + \frac{π}{6}$

substituting $v = \frac{y}{x}$ M1

Note: This M1 may be awarded earlier.

$y = x (\sqrt{3} tan (\sqrt{3} ln x + \frac{π}{6}) + 1)$ A1

[10 marks]

c.ii.

curve drawn over correct domain A1

[1 mark]

c.iii.

the sketch shows that $f$ is concave up A1

Note: Accept $f^{'}$ is increasing.

this means the tangent drawn using Euler’s method will give an underestimate of the real value, so $f (1.4)$ > estimate in part (a) R1

Note: The R1 is dependent on the A1.

[2 marks]

c.iv.

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

c.iii.

[N/A]

c.iv.

20N.3.AHL.TZ0.Hca_1

Question

Use l’Hôpital’s rule to find

$\lim_{x \to 1} \frac{\cos (x^{2} - 1) - 1}{e^{x - 1} - x}$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to use l’Hôpital’s rule M1

$= \lim_{x \to 1} \frac{- 2 x \sin (x^{2} - 1)}{e^{x - 1} - 1}$ A1A1

Note: Award A1 for the numerator and A1 for the denominator.

substitution of $1$ into their expression (M1)

$= \frac{0}{0}$ hence use l’Hôpital’s rule again

Note: If the first use of l’Hôpital’s rule results in an expression which is not in indeterminate form, do not award any further marks.

attempt to use product rule in numerator M1

$= \lim_{x \to 1} \frac{- 4 x^{2} \cos (x^{2} - 1) - 2 \sin (x^{2} - 1)}{e^{x - 1}}$ A1

$= - 4$ A1

[7 marks]

Examiners report

[N/A]

20N.3.AHL.TZ0.Hca_3b

Question

The curve $y = f (x)$ has a gradient function given by

$\frac{d y}{d x} = x - y$ .

The curve passes through the point $(1, 1)$ .

On the same set of axes, sketch and label isoclines for $\frac{d y}{d x} = - 1, 0$ and $1$ , and clearly indicate the value of each $y$ -intercept.

[3]

a.i.

Hence or otherwise, explain why the point $(1, 1)$ is a local minimum.

[3]

a.ii.

Find the solution of the differential equation $\frac{d y}{d x} = x - y$ , which passes through the point $(1, 1)$ . Give your answer in the form $y = f (x)$ .

[8]

Explain why the graph of $y = f (x)$ does not intersect the isocline $\frac{d y}{d x} = 1$ .

[2]

c.i.

Sketch the graph of $y = f (x)$ on the same set of axes as part (a)(i).

[2]

c.ii.

Markscheme

attempt to find equation of isoclines by setting $x - y = - 1, 0, 1$ M1

$3$ parallel lines with positive gradient A1

$y$ -intercept $= - c$ for $\frac{d y}{d x} = c$ A1

Note: To award A1, each $y$ -intercept should be clear, but condone a missing label (eg. $(0, 0)$ ).

If candidates represent the lines using slope fields, but omit the lines, award maximum of M1A0A1.

[3 marks]

a.i.

at point $(1, 1), \frac{d y}{d x} = 0$ A1

EITHER

to the left of $(1, 1)$ , the gradient is negative R1

to the right of $(1, 1)$ , the gradient is positive R1

Note: Accept any correct reasoning using gradient, isoclines or slope field.

If a candidate uses left/right or $x < 1 / x > 1$ without explicitly referring to the point $(1, 1)$ or a correct region on the diagram, award R0R1.

$\frac{d^{2} y}{d x^{2}} = 1 - \frac{d y}{d x}$ A1

$\frac{d^{2} y}{d x^{2}} = 1 (> 0)$ A1

Note: accept correct reasoning $\frac{d y}{d x}$ that is increasing as $x$ increases.

THEN

hence $(1, 1)$ is a local minimum AG

[3 marks]

a.ii.

integrating factor $= e^{\int d x}$ (M1)

$= e^{x}$ (A1)

$\frac{d y}{d x} e^{x} + y e^{x} = x e^{x}$ (M1)

$y e^{x} = \int x e^{x} d x$ A1

$= x e^{x} - \int e^{x} d x$ (M1)

$= x e^{x} - e^{x} (+ c)$ A1

Note: Award A1 for the correct RHS.

substituting $(1, 1)$ gives

$e = e - e + c$ M1

$c = e$

$y = x - 1 + e^{1 - x}$ A1

[8 marks]

METHOD 1

EITHER

attempt to solve for the intersection $x - 1 + e^{1 - x} = x - 1$ (M1)

attempt to find the difference $x - 1 + e^{1 - x} - (x - 1)$ (M1)

THEN

$e^{1 - x} > 0$ for all $x$ R1

Note: Accept $e^{1 - x} \neq 0$ or equivalent reasoning.

therefore the curve does not intersect the isocline AG

METHOD 2

$y = x - 1$ is an (oblique) asymptote to the curve R1

Note: Do not accept “the curve is parallel to $y = x - 1$ "

$y = x - 1$ is the isocline for $\frac{d y}{d x} = 1$ R1

therefore the curve does not intersect the isocline AG

METHOD 3

The initial point is above $y = x - 1$ , so $\frac{d y}{d x} < 1$ R1

$\Rightarrow x - y < 1$

$\Rightarrow y > x - 1$ R1

therefore the curve does not intersect the isocline AG

[2 marks]

c.i.

concave up curve with minimum at approximately $(1, 1)$ A1

asymptote of curve is isocline $y = x - 1$ A1

Note: Only award FT from (b) if the above conditions are satisfied.

[2 marks]

c.ii.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

c.i.

[N/A]

c.ii.

20N.3.AHL.TZ0.Hca_4b

Question

The function $f$ is defined by $f (x) = \ln (1 + x^{2})$ where $- 1 < x < 1$ .

The seventh derivative of $f$ is given by $f^{(7)} (x) = \frac{1440 x (x^{6} - 21 x^{4} + 35 x^{2} - 7)}{{(1 + x^{2})}^{7}}$ .

Use the Maclaurin series for $\ln (1 + x)$ to write down the first three non-zero terms of the Maclaurin series for $f (x)$ .

[2]

a.i.

Hence find the first three non-zero terms of the Maclaurin series for $\frac{x}{1 + x^{2}}$ .

[4]

a.ii.

Use your answer to part (a)(i) to write down an estimate for $f (0.4)$ .

[1]

Use the Lagrange form of the error term to find an upper bound for the absolute value of the error in calculating $f (0.4)$ , using the first three non-zero terms of the Maclaurin series for $f (x)$ .

[6]

c.i.

With reference to the Lagrange form of the error term, explain whether your answer to part (b) is an overestimate or an underestimate for $f (0.4)$ .

[2]

c.ii.

Markscheme

substitution of $x^{2}$ in $\ln (1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \dots$ (M1)

$x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{3}$ A1

[2 marks]

a.i.

$\frac{d}{d x} (\ln (1 + x^{2})) = \frac{2 x}{1 + x^{2}}$ (M1)

Note: Award (M1) if this is seen in part (a)(i).

attempt to differentiate their answer in part (a) (M1)

$\frac{2 x}{1 + x^{2}} = 2 x - \frac{4 x^{3}}{2} + \frac{6 x^{5}}{3}$ M1

Note: Award M1 for equating their derivatives.

$\frac{x}{1 + x^{2}} = x - x^{3} + x^{5}$ A1

[4 marks]

a.ii.

$f (0.4) \approx 0.149$ A1

Note: Accept an answer that rounds correct to $2s.f.$ or better.

[1 mark]

attempt to find the maximum of $|f^{(7)} (c)|$ for $c \in [0, 0.4]$ (M1)

maximum of $|f^{(7)} (c)|$ occurs at $c = 0.199$ (A1)

$|f^{(7)} (c)| < 1232.97 \dots$ (for all $c \in] 0, 0.4 [$ ) (A1)

use of $x = 0.4$ (M1)

substitution of $n = 6$ and $a = 0$ and their value of $x$ and their value of $f^{(7)} (c)$ into Lagrange error term (M1)

Note: Award (M1) for substitution of $n = 3$ and $a = 0$ and their value of $x$ and their value of $f^{(4)} (c)$ into Lagrange error term.

$|R_{6} (0.4)| < \frac{1232.97 {(0.4)}^{7}}{7!}$

upper bound $= 0.000401$ A1

Note: Accept an answer that rounds correct to $1s.f$ or better.

[6 marks]

c.i.

$f^{(7)} (c) < 0$ (for all $c \in] 0, 0.4 [$ ) R1

Note: Accept $R_{6} (c) < 0$ or “the error term is negative”.

the answer in (b) is an overestimate A1

Note: The A1 is dependent on the R1.

[2 marks]

c.ii.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

c.i.

[N/A]

c.ii.

21M.3.AHL.TZ1.1b

Question

This question asks you to explore the behaviour and key features of cubic polynomials of the form $x^{3} - 3 c x + d$ .

Consider the function $f (x) = x^{3} - 3 c x + 2$ for $x \in ℝ$ and where $c$ is a parameter, $c \in ℝ$ .

The graphs of $y = f (x)$ for $c = - 1$ and $c = 0$ are shown in the following diagrams.

$c = - 1$ $c = 0$

On separate axes, sketch the graph of $y = f (x)$ showing the value of the $y$ -intercept and the coordinates of any points with zero gradient, for

Hence, or otherwise, find the set of values of $c$ such that the graph of $y = f (x)$ has

Given that the graph of $y = f (x)$ has one local maximum point and one local minimum point, show that

Hence, for $c > 0$ , find the set of values of $c$ such that the graph of $y = f (x)$ has

$c = 1$ .

[3]

a.i.

$c = 2$ .

[3]

a.ii.

Write down an expression for $f' (x)$ .

[1]

a point of inflexion with zero gradient.

[1]

c.i.

one local maximum point and one local minimum point.

[2]

c.ii.

no points where the gradient is equal to zero.

[1]

c.iii.

the $y$ -coordinate of the local maximum point is $2 c^{\frac{3}{2}} + 2$ .

[3]

d.i.

the $y$ -coordinate of the local minimum point is $- 2 c^{\frac{3}{2}} + 2$ .

[1]

d.ii.

exactly one $x$ -axis intercept.

[2]

e.i.

exactly two $x$ -axis intercepts.

[2]

e.ii.

exactly three $x$ -axis intercepts.

[2]

e.iii.

Consider the function $g (x) = x^{3} - 3 c x + d$ for $x \in ℝ$ and where $c, d \in ℝ$ .

Find all conditions on $c$ and $d$ such that the graph of $y = g (x)$ has exactly one $x$ -axis intercept, explaining your reasoning.

[6]

Markscheme

$c = 1$ : positive cubic with correct $y$ -intercept labelled A1

local maximum point correctly labelled A1

local minimum point correctly labelled A1

[3 marks]

a.i.

$c = 2$ : positive cubic with correct $y$ -intercept labelled A1

local maximum point correctly labelled A1

local minimum point correctly labelled A1

Note: Accept the following exact answers:
Local maximum point coordinates $(- \sqrt{2}, 2 + 4 \sqrt{2})$ .
Local minimum point coordinates $(\sqrt{2}, 2 - 4 \sqrt{2})$ .

[3 marks]

a.ii.

$f' (x) = 3 x^{2} - 3 c$ A1

Note: Accept $3 x^{2} - 3 c$ (an expression).

[1 mark]

$c = 0$ A1

[1 mark]

c.i.

considers the number of solutions to their $f' (x) = 0$ (M1)

$3 x^{2} - 3 c = 0$

$c > 0$ A1

[2 marks]

c.ii.

$c < 0$ A1

Note: The (M1) in part (c)(ii) can be awarded for work shown in either (ii) or (iii).

[1 mark]

c.iii.

attempts to solve their $f' (x) = 0$ for $x$ (M1)

$x \pm \sqrt{c}$ (A1)

Note: Award (A1) if either $x = - \sqrt{c}$ or $x = \sqrt{c}$ is subsequently considered.
Award the above (M1)(A1) if this work is seen in part (c).

correctly evaluates $f (- \sqrt{c})$ A1

$f (- \sqrt{c}) = - c^{\frac{3}{2}} + 3 c^{\frac{3}{2}} + 2 (= - c \sqrt{c} + 3 c \sqrt{c} + 2)$

the $y$ -coordinate of the local maximum point is $2 c^{\frac{3}{2}} + 2$ AG

[3 marks]

d.i.

correctly evaluates $f (\sqrt{c})$ A1

$f (\sqrt{c}) = c^{\frac{3}{2}} - 3 c^{\frac{3}{2}} + 2 (= c \sqrt{c} - 3 c \sqrt{c} + 2)$

the $y$ -coordinate of the local minimum point is $- 2 c^{\frac{3}{2}} + 2$ AG

[1 mark]

d.ii.

the graph of $y = f (x)$ will have one $x$ -axis intercept if

EITHER

$- 2 c^{\frac{3}{2}} + 2 > 0$ (or equivalent reasoning) R1

the minimum point is above the $x$ -axis R1

Note: Award R1 for a rigorous approach that does not (only) refer to sketched graphs.

THEN

$0 < c < 1$ A1

Note: Condone $c < 1$ . The A1 is independent of the R1.

[2 marks]

e.i.

the graph of $y = f (x)$ will have two $x$ -axis intercepts if

EITHER

$- 2 c^{\frac{3}{2}} + 2 = 0$ (or equivalent reasoning) (M1)

evidence from the graph in part(a)(i) (M1)

THEN

$c = 1$ A1

[2 marks]

e.ii.

the graph of $y = f (x)$ will have three $x$ -axis intercepts if

EITHER

$- 2 c^{\frac{3}{2}} + 2 < 0$ (or equivalent reasoning) (M1)

reasoning from the results in both parts (e)(i) and (e)(ii) (M1)

THEN

$c > 1$ A1

[2 marks]

e.iii.

case 1:

$c \leq 0$ (independent of the value of $d$ ) A1

EITHER

$g' (x) = 0$ does not have two solutions (has no solutions or $1$ solution) R1

$\Rightarrow g' (x) \geq 0$ for $x \in^{~}$ R1

the graph of $y = f (x)$ has no local maximum or local minimum points, hence any vertical translation of this graph ( $y = g (x)$ ) will also have no local maximum or local minimum points R1

THEN

therefore there is only one $x$ -axis intercept AG

Note: Award at most A0R1 if only $c < 0$ is considered.

case 2

$c > 0$

$(- \sqrt{c}, 2 c^{\frac{3}{2}} + d)$ is a local maximum point and $(\sqrt{c}, - 2 c^{\frac{3}{2}} + d)$ is a local minimum point (A1)

Note: Award (A1) for a correct $y$ -coordinate seen for either the maximum or the minimum.

considers the positions of the local maximum point and/or the local minimum point (M1)

EITHER
considers both points above the $x$ -axis or both points below the $x$ -axis

OR

considers either the local minimum point only above the $x$ -axis OR the local maximum point only below the $x$ -axis

THEN

$d > 2 c^{\frac{3}{2}}$ (both points above the $x$ -axis) A1

$d < - 2 c^{\frac{3}{2}}$ (both points above the $x$ -axis) A1

Note: Award at most (A1)(M1)A0A0 for case 2 if $c > 0$ is not clearly stated.

[6 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

c.iii.

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

e.i.

[N/A]

e.ii.

[N/A]

e.iii.

[N/A]

21M.3.AHL.TZ2.1c

Question

This question asks you to explore the behaviour and some key features of the function $f_{n} (x) = x^{n} (a - x)^{n}$ , where $a \in ℝ^{+}$ and $n \in ℤ^{+}$ .

In parts (a) and (b), only consider the case where $a = 2$ .

Consider $f_{1} (x) = x (2 - x)$ .

Consider $f_{n} (x) = x^{n} {(2 - x)}^{n}$ , where $n \in ℤ^{+}, n > 1$ .

Now consider $f_{n} (x) = x^{n} {(a - x)}^{n}$ where $a \in ℝ^{+}$ and $n \in ℤ^{+}, n > 1$ .

By using the result from part (f) and considering the sign of ${f_{n}}^{'} (- 1)$ , show that the point $(0, 0)$ on the graph of $y = f_{n} (x)$ is

Sketch the graph of $y = f_{1} (x)$ , stating the values of any axes intercepts and the coordinates of any local maximum or minimum points.

[3]

Use your graphic display calculator to explore the graph of $y = f_{n} (x)$ for

• the odd values $n = 3$ and $n = 5$ ;

• the even values $n = 2$ and $n = 4$ .

Hence, copy and complete the following table.

[6]

Show that ${f_{n}}^{'} (x) = n x^{n - 1} (a - 2 x) {(a - x)}^{n - 1}$ .

[5]

State the three solutions to the equation ${f_{n}}^{'} (x) = 0$ .

[2]

Show that the point $(\frac{a}{2}, f_{n} (\frac{a}{2}))$ on the graph of $y = f_{n} (x)$ is always above the horizontal axis.

[3]

Hence, or otherwise, show that ${f_{n}}^{'} (\frac{a}{4}) > 0$ , for $n \in ℤ^{+}$ .

[2]

a local minimum point for even values of $n$ , where $n > 1$ and $a \in ℝ^{+}$ .

[3]

g.i.

a point of inflexion with zero gradient for odd values of $n$ , where $n > 1$ and $a \in ℝ^{+}$ .

[2]

g.ii.

Consider the graph of $y = x^{n} {(a - x)}^{n} - k$ , where $n \in ℤ^{+}$ , $a \in ℝ^{+}$ and $k \in ℝ$ .

State the conditions on $n$ and $k$ such that the equation $x^{n} {(a - x)}^{n} = k$ has four solutions for $x$ .

[5]

Markscheme

inverted parabola extended below the $x$ -axis A1

$x$ -axis intercept values $x = 0, 2$ A1

Note: Accept a graph passing through the origin as an indication of $x = 0$ .

local maximum at $(1, 1)$ A1

Note: Coordinates must be stated to gain the final A1.
Do not accept decimal approximations.

[3 marks]

A1A1A1A1A1A1

Note: Award A1 for each correct value.

For a table not sufficiently or clearly labelled, assume that their values are in the same order as the table in the question paper and award marks accordingly.

[6 marks]

METHOD 1

attempts to use the product rule (M1)

${f_{n}}^{'} (x) = - n x^{n} {(a - x)}^{n - 1} + n x^{n - 1} {(a - x)}^{n}$ A1A1

Note: Award A1 for a correct $u \frac{d v}{d x}$ and A1 for a correct $v \frac{d u}{d x}$ .

EITHER

attempts to factorise ${f_{n}}^{'} (x)$ (involving at least one of $n x^{n - 1}$ or ${(a - x)}^{n - 1}$ ) (M1)

$= n x^{n - 1} {(a - x)}^{n - 1} ((a - x) - x)$ A1

attempts to express ${f_{n}}^{'} (x)$ as the difference of two products with each product containing at least one of $n x^{n - 1}$ or ${(a - x)}^{n - 1}$ (M1)

$= (- x) (n x^{n - 1}) {(a - x)}^{n - 1} + (a - x) (n x^{n - 1}) {(a - x)}^{n - 1}$ A1

THEN

${f_{n}}^{'} (x) = n x^{n - 1} (a - 2 x) {(a - x)}^{n - 1}$ AG

Note: Award the final (M1)A1 for obtaining any of the following forms:

${f_{n}}^{'} (x) = n x^{n} {(a - x)}^{n} (\frac{a - x - x}{x (a - x)}); {f_{n}}^{'} (x) = \frac{n x^{n} {(a - x)}^{n}}{x (a - x)} (a - x - x);$

${f_{n}}^{'} (x) = n x^{n - 1} ({(a - x)}^{n} - x {(a - x)}^{n - 1});$

${f_{n}}^{'} (x) = {(a - x)}^{n - 1} (n x^{n - 1} {(a - x)}^{n} - n x^{n})$

METHOD 2

$f_{n} (x) = {(x (a - x))}^{n}$ (M1)

$= {(a x - x^{2})}^{n}$ A1

attempts to use the chain rule (M1)

${f_{n}}^{'} (x) = n (a - 2 x) {(a x - x^{2})}^{n - 1}$ A1A1

Note: Award A1 for $n (a - 2 x)$ and A1 for ${(a x - x^{2})}^{n - 1}$ .

${f_{n}}^{'} (x) = n x^{n - 1} (a - 2 x) {(a - x)}^{n - 1}$ AG

[5 marks]

$x = 0, x = \frac{a}{2}, x = a$ A2

Note: Award A1 for either two correct solutions or for obtaining $x = 0, x = - a, x = - \frac{a}{2}$
Award A0 otherwise.

[2 marks]

attempts to find an expression for $f_{n} (\frac{a}{2})$ (M1)

$f_{n} (\frac{a}{2}) = {(\frac{a}{2})}^{n} {(a - \frac{a}{2})}^{n}$

$= {(\frac{a}{2})}^{n} {(\frac{a}{2})}^{n} (= {(\frac{a}{2})}^{2 n}), (= {({(\frac{a}{2})}^{n})}^{2})$ A1

EITHER

since $a \in ℝ^{+}, {(\frac{a}{2})}^{2 n} > 0$ (for $n \in ℤ^{+}, n > 1$ and so $f_{n} (\frac{a}{2}) > 0$ ) R1

Note: Accept any logically equivalent conditions/statements on $a$ and $n$ .
Award R0 if any conditions/statements specified involving $a$ , $n$ or both are incorrect.

(since $a \in ℝ^{+}$ ), $\frac{a}{2}$ raised to an even power ( $2 n$ ) (or equivalent reasoning) is always positive (and so $f_{n} (\frac{a}{2}) > 0$ ) R1

Note: The condition $a \in ℝ^{+}$ is given in the question. Hence some candidates will assume $a \in ℝ^{+}$ and not state it. In these instances, award R1 for a convincing argument.
Accept any logically equivalent conditions/statements on on $a$ and $n$ .
Award R0 if any conditions/statements specified involving $a$ , $n$ or both are incorrect.

THEN

so $(\frac{a}{2}, f_{n} (\frac{a}{2}))$ is always above the horizontal axis AG

Note: Do not award (M1)A0R1.

[3 marks]

METHOD 1

${f_{n}}^{'} (\frac{a}{4}) = n {(\frac{a}{4})}^{n - 1} (a - \frac{a}{2}) {(a - \frac{a}{4})}^{n - 1} (= n {(\frac{a}{4})}^{n - 1} (\frac{a}{2}) {(\frac{3 a}{4})}^{n - 1})$ A1

EITHER

$n {(\frac{a}{4})}^{n - 1} (\frac{a}{2}) {(\frac{3 a}{4})}^{n - 1} > 0$ as $a \in ℝ^{+}$ and $n \in ℤ^{+}$ R1

$n {(\frac{a}{4})}^{n - 1}, (a - \frac{a}{2})$ and ${(a - \frac{a}{4})}^{n - 1}$ are all $> 0$ R1

Note: Do not award A0R1.
Accept equivalent reasoning on correct alternative expressions for ${f_{n}}^{'} (\frac{a}{4})$ and accept any logically equivalent conditions/statements on $a$ and $n$ .

Exceptions to the above are condone $n > 1$ and condone $n > 0$ .

An alternative form for ${f_{n}}^{'} (\frac{a}{4})$ is $(2 n) {(3)}^{n - 1} {(\frac{a}{4})}^{2 n - 1}$ .

THEN

hence ${f_{n}}^{'} (\frac{a}{4}) > 0$ AG

METHOD 2

$f_{n} (0) = 0$ and $f_{n} (\frac{a}{2}) > 0$ A1

(since $f_{n}$ is continuous and there are no stationary points between $x = 0$ and $x = \frac{a}{2}$ )

the gradient (of the curve) must be positive between $x = 0$ and $x = \frac{a}{2}$ R1

Note: Do not award A0R1.

hence ${f_{n}}^{'} (\frac{a}{4}) > 0$ AG

[2 marks]

${f_{n}}^{'} (- 1) = n {(- 1)}^{n - 1} (a + 2) {(a + 1)}^{n - 1}$

for $n$ even:

$n {(- 1)}^{n - 1} (= - n) < 0$ (and $(a + 2), {(a + 1)}^{n - 1}$ are both $> 0$ ) R1

${f_{n}}^{'} (- 1) < 0$ A1

${f_{n}}^{'} (0) = 0$ and ${f_{n}}^{'} (\frac{a}{4}) > 0$ (seen anywhere) A1

Note: Candidates can give arguments based on the sign of ${(- 1)}^{n - 1}$ to obtain the R mark.
For example, award R1 for the following:
If $n$ is even, then $n - 1$ is odd and hence ${(- 1)}^{n - 1} < 0 (= - 1)$ .
Do not award R0A1.
The second A1 is independent of the other two marks.
The A marks can be awarded for correct descriptions expressed in words.
Candidates can state $(0, 0)$ as a point of zero gradient from part (d) or show, state or explain (words or diagram) that ${f_{n}}^{'} (0) = 0$ . The last A mark can be awarded for a clearly labelled diagram showing changes in the sign of the gradient.
The last A1 can be awarded for use of a specific case (e.g. $n = 2$ ).

hence $(0, 0)$ is a local minimum point AG

[3 marks]

g.i.

for $n$ odd:

$n {(- 1)}^{n - 1} (= n) < 0$ , (and $(a + 2), {(a + 1)}^{n - 1}$ are both $> 0$ ) so ${f_{n}}^{'} (- 1) > 0$ R1

Note: Candidates can give arguments based on the sign of ${(- 1)}^{n - 1}$ to obtain the R mark.
For example, award R1 for the following:
If $n$ is odd, then $n - 1$ is even and hence ${(- 1)}^{n - 1} > 0 (= 1)$ .

${f_{n}}^{'} (0) = 0$ and ${f_{n}}^{'} (\frac{a}{4}) > 0$ (seen anywhere) A1

Note: The A1 is independent of the R1.
Candidates can state $(0, 0)$ as a point of zero gradient from part (d) or show, state or explain (words or diagram) that ${f_{n}}^{'} (0) = 0$ . The last A mark can be awarded for a clearly labelled diagram showing changes in the sign of the gradient.
The last A1 can be awarded for use of a specific case (e.g. $n = 3$ ).

hence $(0, 0)$ is a point of inflexion with zero gradient AG

[2 marks]

g.ii.

considers the parity of $n$ (M1)

Note: Award M1 for stating at least one specific even value of $n$ .

$n$ must be even (for four solutions) A1

Note: The above 2 marks are independent of the 3 marks below.

$0 < k < {(\frac{a}{2})}^{2 n}$ A1A1A1

Note: Award A1 for the correct lower endpoint, A1 for the correct upper endpoint and A1 for strict inequality signs.

The third A1 (strict inequality signs) can only be awarded if A1A1 has been awarded.
For example, award A1A1A0 for $0 \leq k \leq {(\frac{a}{2})}^{2 n}$ . Award A1A0A0 for $k > 0$ .

Award A1A0A0 for $0 < k < f_{n} (\frac{a}{2})$ .

[5 marks]

Examiners report

[N/A]

g.i.

[N/A]

g.ii.

[N/A]

21M.3.AHL.TZ1.2e.i

Question

This question asks you to examine various polygons for which the numerical value of the area is the same as the numerical value of the perimeter. For example, a $3$ by $6$ rectangle has an area of $18$ and a perimeter of $18$ .

For each polygon in this question, let the numerical value of its area be $A$ and let the numerical value of its perimeter be $P$ .

An $n$ -sided regular polygon can be divided into $n$ congruent isosceles triangles. Let $x$ be the length of each of the two equal sides of one such isosceles triangle and let $y$ be the length of the third side. The included angle between the two equal sides has magnitude $\frac{2 π}{n}$ .

Part of such an $n$ -sided regular polygon is shown in the following diagram.

Consider a $n$ -sided regular polygon such that $A = P$ .

The Maclaurin series for $\tan x$ is $x + \frac{x^{3}}{3} + \frac{2 x^{5}}{15} + \dots$

Consider a right-angled triangle with side lengths $a, b$ and $\sqrt{a^{2} + b^{2}}$ , where $a \geq b$ , such that $A = P$ .

Find the side length, $s$ , where $s > 0$ , of a square such that $A = P$ .

[3]

Write down, in terms of $x$ and $n$ , an expression for the area, $A_{T}$ , of one of these isosceles triangles.

[1]

Show that $y = 2 x \sin \frac{π}{n}$ .

[2]

Use the results from parts (b) and (c) to show that $A = P = 4 n \tan \frac{π}{n}$ .

[7]

Use the Maclaurin series for $\tan x$ to find $\lim_{n \to \infty} (4 n \tan \frac{π}{n})$ .

[3]

e.i.

Interpret your answer to part (e)(i) geometrically.

[1]

e.ii.

Show that $a = \frac{8}{b - 4} + 4$ .

[7]

By using the result of part (f) or otherwise, determine the three side lengths of the only two right-angled triangles for which $a, b, A, P \in ℤ$ .

[3]

g.i.

Determine the area and perimeter of these two right-angled triangles.

[1]

g.ii.

Markscheme

$A = s^{2}$ and $P = 4 s$ (A1)

$A = P \Rightarrow s^{2} = 4 s$ (M1)

$s (s - 4) = 0$

$\Rightarrow s = 4 (s > 0)$ A1

Note: Award A1M1A0 if both $s = 4$ and $s = 0$ are stated as final answers.

[3 marks]

$A_{T} = \frac{1}{2} x^{2} \sin \frac{2 π}{n}$ A1

Note: Award A1 for a correct alternative form expressed in terms of $x$ and $n$ only.

For example, using Pythagoras’ theorem, $A_{T} = x \sin \frac{π}{n} \sqrt{x^{2} - x^{2} \sin^{2} \frac{π}{n}}$ or $A_{T} = 2 (\frac{1}{2} (x \sin \frac{π}{n}) (x \cos \frac{π}{n}))$ or $A_{T} = x^{2} \sin \frac{π}{n} \cos \frac{π}{n}$ .

[1 mark]

METHOD 1

uses $\sin θ = \frac{opp}{hyp}$ (M1)

$\frac{\frac{y}{2}}{x} = \sin \frac{π}{n}$ A1

$y = 2 x \sin \frac{π}{n}$ AG

METHOD 2

uses Pythagoras’ theorem ${(\frac{y}{2})}^{2} + h^{2} = x^{2}$ and $h = x \cos \frac{π}{n}$ (M1)

${(\frac{y}{2})}^{2} + {(x \cos \frac{π}{n})}^{2} = x^{2} (y^{2} = 4 x^{2} (1 - \cos^{2} \frac{π}{n}))$

$= 4 x^{2} \sin^{2} \frac{π}{n}$ A1

$y = 2 x \sin \frac{π}{n}$ AG

METHOD 3

uses the cosine rule (M1)

$y^{2} = 2 x^{2} - 2 x^{2} \cos \frac{2 π}{n} (= 2 x^{2} (1 - \cos \frac{2 π}{n}))$

$= 4 x^{2} \sin^{2} \frac{π}{n}$ A1

$y = 2 x \sin \frac{π}{n}$ AG

METHOD 4

uses the sine rule (M1)

$\frac{y}{\sin \frac{2 π}{n}} = \frac{x}{\sin (\frac{π}{2} - \frac{π}{n})}$

$y \cos \frac{π}{n} = 2 x \sin \frac{π}{n} \cos \frac{π}{n}$ A1

$y = 2 x \sin \frac{π}{n}$ AG

[2 marks]

$A = P \Rightarrow n A_{T} = n y$ (M1)

Note: Award M1 for equating correct expressions for $A$ and $P$ .

$\frac{1}{2} n x^{2} \sin \frac{2 π}{n} = 2 n x \sin \frac{π}{n} (n x^{2} \sin \frac{π}{n} \cos \frac{π}{n} = 2 n x \sin \frac{π}{n})$

$\frac{1}{2} x^{2} \sin \frac{2 π}{n} = 2 x \sin \frac{π}{n} (x^{2} \sin \frac{π}{n} \cos \frac{π}{n} = 2 x \sin \frac{π}{n})$ A1

uses $\sin \frac{2 π}{n} = 2 \sin \frac{π}{n} \cos \frac{π}{n}$ (seen anywhere in part (d) or in part (b)) (M1)

$x^{2} \sin \frac{π}{n} \cos \frac{π}{n} = 2 x \sin \frac{π}{n}$

attempts to either factorise or divide their expression (M1)

$x \sin \frac{π}{n} (x \cos \frac{π}{n} - 2) = 0$

$x = \frac{2}{\cos \frac{π}{n}}, (x \sin \frac{π}{n} \neq 0)$ (or equivalent) A1

EITHER

substitutes $x = \frac{2}{\cos \frac{π}{n}}$ (or equivalent) into $P = n y$ (M1)

$P = 2 n (\frac{2}{\cos \frac{π}{n}}) (\sin \frac{π}{n})$ A1

Note: Other approaches are possible. For example, award A1 for $P = 2 n x \cos \frac{π}{n} \tan \frac{π}{n}$ and M1 for substituting $x = \frac{2}{\cos \frac{π}{n}}$ into $P$ .

substitutes $x = \frac{2}{\cos \frac{π}{n}}$ (or equivalent) into $A = n A_{T}$ (M1)

$A = \frac{1}{2} n {(\frac{2}{\cos \frac{π}{n}})}^{2} (\sin \frac{2 π}{n})$

$A = \frac{1}{2} n {(\frac{2}{\cos \frac{π}{n}})}^{2} (2 \sin \frac{π}{n} \cos \frac{π}{n})$ A1

THEN

$A = P = 4 n \tan \frac{π}{n}$ AG

[7 marks]

attempts to use the Maclaurin series for $\tan x$ with $x = \frac{π}{n}$ (M1)

$\tan \frac{π}{n} = \frac{π}{n} + \frac{{(\frac{π}{n})}^{3}}{3} + \frac{2 {(\frac{π}{n})}^{5}}{15} (+ \dots)$

$4 n \tan \frac{π}{n} = 4 n (\frac{π}{n} + \frac{π^{3}}{3 n^{3}} + \frac{2 π^{5}}{15 n^{5}} (+ \dots))$ (or equivalent) A1

$= 4 (π + \frac{π^{3}}{3 n^{2}} + \frac{2 π^{5}}{15 n^{4}} + \dots)$

$\Rightarrow \lim_{n \to \infty} (4 n \tan \frac{π}{n}) = 4 π$ A1

Note: Award a maximum of M1A1A0 if $\lim_{n \to \infty}$ is not stated anywhere.

[3 marks]

e.i.

(as $n \to \infty, P \to 4 π$ and $A \to 4 π$ )

the polygon becomes a circle of radius $2$ R1

Note: Award R1 for alternative responses such as:
the polygon becomes a circle of area $4 π$ OR
the polygon becomes a circle of perimeter $4 π$ OR
the polygon becomes a circle with $A = P = 4 π$ .
Award R0 for polygon becomes a circle.

[1 mark]

e.ii.

$A = \frac{1}{2} a b$ and $P = a + b + \sqrt{a^{2} + b^{2}}$ (A1)(A1)

equates their expressions for $A$ and $P$ M1

$A = P \Rightarrow a + b + \sqrt{a^{2} + b^{2}} = \frac{1}{2} a b$

$\sqrt{a^{2} + b^{2}} = \frac{1}{2} a b - (a + b)$ M1

Note: Award M1 for isolating $\sqrt{a^{2} + b^{2}}$ or $\pm 2 \sqrt{a^{2} + b^{2}}$ . This step may be seen later.

$a^{2} + b^{2} = {(\frac{1}{2} a b - (a + b))}^{2}$

$a^{2} + b^{2} = \frac{1}{4} a^{2} b^{2} - 2 (\frac{1}{2} a b) (a + b) + {(a + b)}^{2}$ M1

$(= \frac{1}{4} a^{2} b^{2} - a^{2} b - a b^{2} + a^{2} + 2 a b + b^{2})$

Note: Award M1 for attempting to expand their RHS of either $a^{2} + b^{2} = \dots$ or $4 (a^{2} + b^{2}) = \dots$ .

EITHER

$a b (\frac{1}{4} a b - a - b + 2) = 0 (a b \neq 0)$ A1

$\frac{1}{4} a b - a - b + 2 = 0$

$a b - 4 a = 4 b - 8$

$\frac{1}{4} a^{2} b^{2} - a^{2} b - a b^{2} + 2 a b = 0$

$a (\frac{1}{4} b^{2} - b) + (2 b - b^{2}) = 0 (a (b^{2} - 4 b) + (8 b - 4 b^{2}) = 0)$ A1

$a = \frac{4 b^{2} - 8 b}{b^{2} - 4 b}$

THEN

$\Rightarrow a = \frac{4 b - 8}{b - 4}$ A1

$a = \frac{4 b - 16 + 8}{b - 4}$

$a = \frac{8}{b - 4} + 4$ AG

Note: Award a maximum of A1A1M1M1M0A0A0 for attempting to verify.
For example, verifying that $A = P = \frac{16}{b - 4} + 2 b + 4$ gains $4$ of the $7$ marks.

[7 marks]

using an appropriate method (M1)

eg substituting values for $b$ or using divisibility properties

$(5, 12, 13)$ and $(6, 8, 10)$ A1A1

Note: Award A1A0 for either one set of three correct side lengths or two sets of two correct side lengths.

[3 marks]

g.i.

$A = P = 30$ and $A = P = 24$ A1

Note: Do not award A1FT.

[1 mark]

g.ii.

Examiners report

[N/A]

e.i.

[N/A]

e.ii.

[N/A]

g.i.

[N/A]

g.ii.

21N.3.AHL.TZ0.1a

Question

In this question you will explore some of the properties of special functions $f$ and $g$ and their relationship with the trigonometric functions, sine and cosine.

Functions $f$ and $g$ are defined as $f (z) = \frac{e^{z} + e^{- z}}{2}$ and $g (z) = \frac{e^{z} - e^{- z}}{2}$ , where $z \in ℂ$ .

Consider $t$ and $u$ , such that $t, u \in ℝ$ .

Using $e^{i u} = \cos u + i \sin u$ , find expressions, in terms of $\sin u$ and $\cos u$ , for

The functions $\cos x$ and $\sin x$ are known as circular functions as the general point ( $\cos θ, \sin θ$ ) defines points on the unit circle with equation $x^{2} + y^{2} = 1$ .

The functions $f (x)$ and $g (x)$ are known as hyperbolic functions, as the general point ( $f (θ), g (θ)$ ) defines points on a curve known as a hyperbola with equation $x^{2} - y^{2} = 1$ . This hyperbola has two asymptotes.

Verify that $u = f (t)$ satisfies the differential equation $\frac{d^{2} u}{d t^{2}} = u$ .

[2]

Show that ${(f (t))}^{2} + {(g (t))}^{2} = f (2 t)$ .

[3]

$f (i u)$ .

[3]

c.i.

$g (i u)$ .

[2]

c.ii.

Hence find, and simplify, an expression for ${(f (i u))}^{2} + {(g (i u))}^{2}$ .

[2]

Show that ${(f (t))}^{2} - {(g (t))}^{2} = {(f (i u))}^{2} - {(g (i u))}^{2}$ .

[4]

Sketch the graph of $x^{2} - y^{2} = 1$ , stating the coordinates of any axis intercepts and the equation of each asymptote.

[4]

The hyperbola with equation $x^{2} - y^{2} = 1$ can be rotated to coincide with the curve defined by $x y = k, k \in ℝ$ .

Find the possible values of $k$ .

[5]

Markscheme

$f' (t) = \frac{e^{t} - e^{- t}}{2}$ A1

$f'' (t) = \frac{e^{t} + e^{- t}}{2}$ A1

$= f (t)$ AG

[2 marks]

METHOD 1

${(f (t))}^{2} + {(g (t))}^{2}$

substituting $f$ and $g$ M1

$= \frac{{(e^{t} + e^{- t})}^{2} + {(e^{t} - e^{- t})}^{2}}{4}$

$= \frac{{(e^{t})}^{2} + 2 + {(e^{- t})}^{2} + {(e^{t})}^{2} - 2 + {(e^{- t})}^{2}}{4}$ (M1)

$= \frac{{(e^{t})}^{2} + {(e^{- t})}^{2}}{2} (= \frac{e^{2 t} + e^{- 2 t}}{2})$ A1

$= f (2 t)$ AG

METHOD 2

$f (2 t) = \frac{e^{2 t} + e^{- 2 t}}{2}$

$= \frac{{(e^{t})}^{2} + {(e^{- t})}^{2}}{2}$ M1

$= \frac{{(e^{t} + e^{- t})}^{2} + {(e^{t} - e^{- t})}^{2}}{4}$ M1A1

$= {(f (t))}^{2} + {(g (t))}^{2}$ AG

Note: Accept combinations of METHODS 1 & 2 that meet at equivalent expressions.

[3 marks]

substituting $e^{i u} = \cos u + i \sin u$ into the expression for $f$ (M1)

obtaining $e^{-i u} = \cos u - i \sin u$ (A1)

$f (i u) = \frac{\cos u + i \sin u + \cos u - i \sin u}{2}$

Note: The M1 can be awarded for the use of sine and cosine being odd and even respectively.

$= \frac{2 \cos u}{2}$

$= \cos u$ A1

[3 marks]

c.i.

$g (i u) = \frac{\cos u + i \sin u - \cos u + i \sin u}{2}$

substituting and attempt to simplify (M1)

$= \frac{2 i \sin u}{2}$

$= i \sin u$ A1

[2 marks]

c.ii.

METHOD 1

${(f (i u))}^{2} + {(g (i u))}^{2}$

substituting expressions found in part (c) (M1)

$= \cos^{2} u - \sin^{2} u (= \cos 2 u)$ A1

METHOD 2

$f (2 i u) = \frac{e^{2 i u} + e^{- 2 i u}}{2}$

$= \frac{\cos 2 u + i \sin 2 u + \cos 2 u - i \sin 2 u}{2}$ M1

$= \cos 2 u$ A1

Note: Accept equivalent final answers that have been simplified removing all imaginary parts eg $2 \cos^{2} u - 1$ etc

[2 marks]

${(f (t))}^{2} - {(g (t))}^{2} = \frac{{(e^{t} + e^{- t})}^{2} - {(e^{t} - e^{- t})}^{2}}{4}$ M1

$= \frac{(e^{2 t} + e^{- 2 t} + 2) - (e^{2 t} + e^{- 2 t} - 2)}{4}$ A1

$= \frac{4}{4} = 1$ A1

Note: Award A1 for a value of 1 obtained from either LHS or RHS of given expression.

${(f (i u))}^{2} - {(g (i u))}^{2} = \cos^{2} u + \sin^{2} u$ M1

$= 1$ (hence ${(f (t))}^{2} - {(g (t))}^{2} = {(f (i u))}^{2} - {(g (i u))}^{2}$ ) AG

Note: Award full marks for showing that ${(f (z))}^{2} - {(g (z))}^{2} = 1, \forall z \in ℂ$ .

[4 marks]

A1A1A1A1

Note: Award A1 for correct curves in the upper quadrants, A1 for correct curves in the lower quadrants, A1 for correct $x$ -intercepts of $(- 1, 0)$ and $(1, 0)$ (condone $x = - 1$ and $1$ ), A1 for $y = x$ and $y = - x$ .

[4 marks]

attempt to rotate by $45 °$ in either direction (M1)

Note: Evidence of an attempt to relate to a sketch of $x y = k$ would be sufficient for this (M1).

attempting to rotate a particular point, eg $(1, 0)$ (M1)

$(1, 0)$ rotates to $(\frac{1}{\sqrt{2}}, \pm \frac{1}{\sqrt{2}})$ (or similar) (A1)

hence $k = \pm \frac{1}{2}$ A1A1

[5 marks]

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

21N.3.AHL.TZ0.2a.i

Question

In this question you will be exploring the strategies required to solve a system of linear differential equations.

Consider the system of linear differential equations of the form:

$\frac{d x}{d t} = x - y$ and $\frac{d y}{d t} = a x + y$ ,

where $x, y, t \in ℝ^{+}$ and $a$ is a parameter.

First consider the case where $a = 0$ .

Now consider the case where $a = - 1$ .

Now consider the case where $a = - 4$ .

From previous cases, we might conjecture that a solution to this differential equation is $y = F e^{λ t}$ , $λ \in ℝ$ and $F$ is a constant.

By solving the differential equation $\frac{d y}{d t} = y$ , show that $y = A e^{t}$ where $A$ is a constant.

[3]

a.i.

Show that $\frac{d x}{d t} - x = - A e^{t}$ .

[1]

a.ii.

Solve the differential equation in part (a)(ii) to find $x$ as a function of $t$ .

[4]

a.iii.

By differentiating $\frac{d y}{d t} = - x + y$ with respect to $t$ , show that $\frac{d^{2} y}{d t^{2}} = 2 \frac{d y}{d t}$ .

[3]

b.i.

By substituting $Y = \frac{d y}{d t}$ , show that $Y = B e^{2 t}$ where $B$ is a constant.

[3]

b.ii.

Hence find $y$ as a function of $t$ .

[2]

b.iii.

Hence show that $x = - \frac{B}{2} e^{2 t} + C$ , where $C$ is a constant.

[3]

b.iv.

Show that $\frac{d^{2} y}{d t^{2}} - 2 \frac{d y}{d t} - 3 y = 0$ .

[3]

c.i.

Find the two values for $λ$ that satisfy $\frac{d^{2} y}{d t^{2}} - 2 \frac{d y}{d t} - 3 y = 0$ .

[4]

c.ii.

Let the two values found in part (c)(ii) be $λ_{1}$ and $λ_{2}$ .

Verify that $y = F e^{λ_{1} t} + G e^{λ_{2} t}$ is a solution to the differential equation in (c)(i),where $G$ is a constant.

[4]

c.iii.

Markscheme

METHOD 1

$\frac{d y}{d t} = y$

$\int \frac{d y}{y} = \int d t$ (M1)

$\ln y = t + c$ OR $\ln |y| = t + c$ A1A1

Note: Award A1 for $\ln y$ and A1 for $t$ and $c$ .

$y = A e^{t}$ AG

METHOD 2

rearranging to $\frac{d y}{d t} - y = 0$ AND multiplying by integrating factor $e^{- t}$ M1

$y e^{- t} = A$ A1A1

$y = A e^{t}$ AG

[3 marks]

a.i.

substituting $y = A e^{t}$ into differential equation in $x$ M1

$\frac{d x}{d t} = x - A e^{t}$

$\frac{d x}{d t} - x = - A e^{t}$ AG

[1 mark]

a.ii.

integrating factor (IF) is $e^{\int - 1 d t}$ (M1)

$= e^{- t}$ (A1)

$e^{- t} \frac{d x}{d t} - x e^{- t} = - A$

$x e^{- t} = - A t + D$ (A1)

$x = (- A t + D) e^{t}$ A1

Note: The first constant must be $A$ , and the second can be any constant for the final A1 to be awarded. Accept a change of constant applied at the end.

[4 marks]

a.iii.

$\frac{d^{2} y}{d t^{2}} = - \frac{d x}{d t} + \frac{d y}{d t}$ A1

EITHER

$= - x + y + \frac{d y}{d t}$ (M1)

$= \frac{d y}{d t} + \frac{d y}{d t}$ A1

$= - x + y + (- x + y)$ (M1)

$= 2 (- x + y)$ A1

THEN

$= 2 \frac{d y}{d t}$ AG

[3 marks]

b.i.

$\frac{d Y}{d t} = 2 Y$ A1

$\int \frac{d Y}{Y} = \int 2 d t$ M1

$\ln |Y| = 2 t + c$ OR $\ln Y = 2 t + c$ A1

$Y = B e^{2 t}$ AG

[3 marks]

b.ii.

$\frac{d y}{d t} = B e^{2 t}$

$y = \int B e^{2 t} d t$ M1

$y = \frac{B}{2} e^{2 t} + C$ A1

Note: The first constant must be $B$ , and the second can be any constant for the final A1 to be awarded. Accept a change of constant applied at the end.

[2 marks]

b.iii.

METHOD 1

substituting $\frac{d y}{d t} = B e^{2 t}$ and their (iii) into $\frac{d y}{d t} = - x + y$ M1(M1)

$B e^{2 t} = - x + \frac{B}{2} e^{2 t} + C$ A1

$x = - \frac{B}{2} e^{2 t} + C$ AG

Note: Follow through from incorrect part (iii) cannot be awarded if it does not lead to the AG.

METHOD 2

$\frac{d x}{d t} = x - \frac{B}{2} e^{2 t} - C$

$\frac{d x}{d t} - x = - \frac{B}{2} e^{2 t} - C$

$\frac{d (x e^{- t})}{d t} = - \frac{B}{2} e^{t} - C e^{- t}$ M1

$x e^{- t} = \int - \frac{B}{2} e^{t} - C e^{- t} d t$

$x e^{- t} = - \frac{B}{2} e^{t} - C e^{- t} + D$ A1

$x = - \frac{B}{2} e^{2 t} + C + D e^{t}$

$\frac{d y}{d t} = - x + y \Rightarrow B e^{2 t} = \frac{B}{2} e^{2 t} - C - D e^{t} + \frac{B}{2} e^{2 t} + C \Rightarrow D = 0$ M1

$x = - \frac{B}{2} e^{2 t} + C$ AG

[3 marks]

b.iv.

$\frac{d y}{d t} = - 4 x + y$

$\frac{d^{2} y}{d t^{2}} = - 4 \frac{d x}{d t} + \frac{d y}{d t}$ seen anywhere M1

METHOD 1

$\frac{d^{2} y}{d t^{2}} = - 4 (x - y) + \frac{d y}{d t}$

attempt to eliminate $x$ M1

$= - 4 (\frac{1}{4} (y - \frac{d y}{d t}) - y) + \frac{d y}{d t}$

$= 2 \frac{d y}{d t} + 3 y$ A1

$\frac{d^{2} y}{d t^{2}} - 2 \frac{d y}{d t} - 3 y = 0$ AG

METHOD 2

rewriting LHS in terms of $x$ and $y$ M1

$\frac{d^{2} y}{d t^{2}} - 2 \frac{d y}{d t} - 3 y = (- 8 x + 5 y) - 2 (- 4 x + y) - 3 y$ A1

$= 0$ AG

[3 marks]

c.i.

$\frac{d y}{d t} = F λ e^{λ t}, \frac{d^{2} y}{d t^{2}} = F λ^{2} e^{λ t}$ (A1)

$F λ^{2} e^{λ t} - 2 F λ e^{λ t} - 3 F e^{λ t} = 0$ (M1)

$λ^{2} - 2 λ - 3 = 0$ (since $e^{λ t} \neq 0$ ) A1

$λ_{1}$ and $λ_{2}$ are $3$ and $- 1$ (either order) A1

[4 marks]

c.ii.

METHOD 1

$y = F e^{3 t} + G e^{- t}$

$\frac{d y}{d t} = 3 F e^{3 t} - G e^{- t}, \frac{d^{2} y}{d t^{2}} = 9 F e^{3 t} - G e^{- t}$ (A1)(A1)

$\frac{d^{2} y}{d t^{2}} - 2 \frac{d y}{d t} - 3 y = 9 F e^{3 t} + G e^{- t} - 2 (3 F e^{3 t} - G e^{- t}) - 3 (F e^{3 t} - G e^{- t})$ M1

$= 9 F e^{3 t} + G e^{- t} - 6 F e^{3 t} + 2 G e^{- t} - 3 F e^{3 t} - 3 G e^{- t}$ A1

$= 0$ AG

METHOD 2

$y = F e^{λ_{1} t} + G e^{λ_{2} t}$

$\frac{d y}{d t} = F λ_{1} e^{λ_{1} t} + G λ_{2} e^{λ_{2} t}, \frac{d^{2} y}{d t^{2}} = F {λ_{1}}^{2} e^{λ_{1} t} + G {λ_{2}}^{2} e^{λ_{2} t}$ (A1)(A1)

$\frac{d^{2} y}{d t^{2}} - 2 \frac{d y}{d t} - 3 y = F {λ_{1}}^{2} e^{λ_{1} t} + G {λ_{2}}^{2} e^{λ_{2} t} - 2 (F λ_{1} e^{λ_{1} t} + G λ_{2} e^{λ_{2} t}) - 3 (F e^{λ_{1} t} + G e^{λ_{2} t})$ M1

$= F e^{λ_{1} t} (λ^{2} - 2 λ - 3) + G e^{λ_{2} t} (λ^{2} - 2 λ - 3)$ A1

$= 0$ AG

[4 marks]

c.iii.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

a.iii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

[N/A]

b.iv.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

c.iii.

22M.3.AHL.TZ2.1e

Question

This question asks you to explore properties of a family of curves of the type $y^{2} = x^{3} + a x + b$ for various values of $a$ and $b$ , where $a, b \in ℕ$ .

On the same set of axes, sketch the following curves for $- 2 \leq x \leq 2$ and $- 2 \leq y \leq 2$ , clearly indicating any points of intersection with the coordinate axes.

Now, consider curves of the form $y^{2} = x^{3} + b$ , for $x \geq - \sqrt[3]{b}$ , where $b \in ℤ^{+}$ .

Next, consider the curve $y^{2} = x^{3} + x, x \geq 0$ .

The curve $y^{2} = x^{3} + x$ has two points of inflexion. Due to the symmetry of the curve these points have the same $x$ -coordinate.

$P (x, y)$ is defined to be a rational point on a curve if $x$ and $y$ are rational numbers.

The tangent to the curve $y^{2} = x^{3} + a x + b$ at a rational point $P$ intersects the curve at another rational point $Q$ .

Let $C$ be the curve $y^{2} = x^{3} + 2$ , for $x \geq - \sqrt[3]{2}$ . The rational point $P (- 1, - 1)$ lies on $C$ .

$y^{2} = x^{3}, x \geq 0$

[2]

a.i.

$y^{2} = x^{3} + 1, x \geq - 1$

[2]

a.ii.

Write down the coordinates of the two points of inflexion on the curve $y^{2} = x^{3} + 1$ .

[1]

b.i.

By considering each curve from part (a), identify two key features that would distinguish one curve from the other.

[1]

b.ii.

By varying the value of $b$ , suggest two key features common to these curves.

[2]

Show that $\frac{d y}{d x} = \pm \frac{3 x^{2} + 1}{2 \sqrt{x^{3} + x}}$ , for $x > 0$ .

[3]

d.i.

Hence deduce that the curve $y^{2} = x^{3} + x$ has no local minimum or maximum points.

[1]

d.ii.

Find the value of this $x$ -coordinate, giving your answer in the form $x = \sqrt{\frac{p \sqrt{3} + q}{r}}$ , where $p, q, r \in ℤ$ .

[7]

Find the equation of the tangent to $C$ at $P$ .

[2]

f.i.

Hence, find the coordinates of the rational point $Q$ where this tangent intersects $C$ , expressing each coordinate as a fraction.

[2]

f.ii.

The point $S (- 1, 1)$ also lies on $C$ . The line $[QS]$ intersects $C$ at a further point. Determine the coordinates of this point.

[5]

Markscheme

approximately symmetric about the $x$ -axis graph of $y^{2} = x^{3}$ A1

including cusp/sharp point at $(0, 0)$ A1

[2 marks]

Note: Final A1 can be awarded if intersections are in approximate correct place with respect to the axes shown. Award A1A1A1A0 if graphs ‘merge’ or ‘cross’ or are discontinuous at $x$ -axis but are otherwise correct. Award A1A0A0A0 if only one correct branch of both curves are seen.

Note: If they sketch graphs on separate axes, award a maximum of 2 marks for the ‘best’ response seen. This is likely to be A1A1A0A0.

a.i.

approximately symmetric about the $x$ -axis graph of $y^{2} = x^{3} + 1$ with approximately correct gradient at axes intercepts A1
some indication of position of intersections at $x = - 1$ , $y = \pm 1$ A1

[2 marks]

Note: If they sketch graphs on separate axes, award a maximum of 2 marks for the ‘best’ response seen. This is likely to be A1A1A0A0.

a.ii.

$(0, 1)$ and $(0, - 1)$ A1

[1 mark]

b.i.

Any two from:

$y^{2} = x^{3}$ has a cusp/sharp point, (the other does not)

graphs have different domains

$y^{2} = x^{3} + 1$ has points of inflexion, (the other does not)

graphs have different $x$ -axis intercepts (one goes through the origin, and the other does not)

graphs have different $y$ -axis intercepts A1

Note: Follow through from their sketch in part (a)(i). In accordance with marking rules, mark their first two responses and ignore any subsequent.

[1 mark]

b.ii.

Any two from:

as , $x \to \infty, y \to \pm \infty$

as $x \to \infty, y^{2} = x^{3} + b$ is approximated by $y^{2} = x^{3}$ (or similar)

they have $x$ intercepts at $x = - \sqrt[3]{b}$

they have $y$ intercepts at $y = (\pm) \sqrt{b}$

they all have the same range

$y = 0$ (or $x$ -axis) is a line of symmetry

they all have the same line of symmetry $(y = 0)$

they have one $x$ -axis intercept

they have two $y$ -axis intercepts

they have two points of inflexion

at $x$ -axis intercepts, curve is vertical/infinite gradient

there is no cusp/sharp point at $x$ -axis intercepts A1A1

Note: The last example is the only valid answer for things “not” present. Do not credit an answer of “they are all symmetrical” without some reference to the line of symmetry.

Note: Do not allow same/ similar shape or equivalent.

Note: In accordance with marking rules, mark their first two responses and ignore any subsequent.

[2 marks]

METHOD 1

attempt to differentiate implicitly M1

$2 y \frac{d y}{d x} = 3 x^{2} + 1$ A1

$\frac{d y}{d x} = \frac{3 x^{2} + 1}{2 y}$ OR $(\pm) 2 \sqrt{x^{3} + x} \frac{d y}{d x} = 3 x^{2} + 1$ A1

$\frac{d y}{d x} = \pm \frac{3 x^{2} + 1}{2 \sqrt{x^{3} + x}}$ AG

METHOD 2

attempt to use chain rule $y = (\pm) \sqrt{x^{3} + x}$ M1

$\frac{d y}{d x} = (\pm) \frac{1}{2} {(x^{3} + x)}^{- \frac{1}{2}} (3 x^{2} + 1)$ A1A1

Note: Award A1 for $(\pm) \frac{1}{2} {(x^{3} + x)}^{- \frac{1}{2}}$ , A1 for $(3 x^{2} + 1)$

$\frac{d y}{d x} = \pm \frac{3 x^{2} + 1}{2 \sqrt{x^{3} + x}}$ AG

[3 marks]

d.i.

EITHER

local minima/maxima occur when $\frac{d y}{d x} = 0$

$1 + 3 x^{2} = 0$ has no (real) solutions (or equivalent) R1

$(x^{2} \geq 0 \Rightarrow) 3 x^{2} + 1 > 0$ , so $\frac{d y}{d x} \neq 0$ R1

THEN

so, no local minima/maxima exist AG

[1 mark]

d.ii.

EITHER

attempt to use quotient rule to find $\frac{d^{2} y}{d x^{2}}$ M1

$\frac{d^{2} y}{d x^{2}} = (\pm) \frac{12 x \sqrt{x + x^{3}} - (1 + 3 x^{2}) {(x + x^{3})}^{- \frac{1}{2}} (1 + 3 x^{2})}{4 (x + x^{3})}$ A1A1

Note: Award A1 for correct $12 x \sqrt{x + x^{3}}$ and correct denominator, A1 for correct $- (1 + 3 x^{2}) {(x + x^{3})}^{- \frac{1}{2}} (1 + 3 x^{2})$ .

Note: Future A marks may be awarded if the denominator is missing or incorrect.

stating or using $\frac{d^{2} y}{d x^{2}} = 0$ (may be seen anywhere) (M1)

$12 x \sqrt{x + x^{3}} = (1 + 3 x^{2}) {(x + x^{3})}^{- \frac{1}{2}} (1 + 3 x^{2})$

attempt to use product rule to find $\frac{d^{2} y}{d x^{2}}$ M1

$\frac{d^{2} y}{d x^{2}} = \frac{1}{2} (3 x^{2} + 1) (- \frac{1}{2}) (3 x^{2} + 1) {(x^{3} + x)}^{- \frac{3}{2}} + 3 x {(x^{3} + x)}^{- \frac{1}{2}}$ A1A1

Note: Award A1 for correct first term, A1 for correct second term.

setting $\frac{d^{2} y}{d x^{2}} = 0$ (M1)

attempts implicit differentiation on $2 y \frac{d y}{d x} = 3 x^{2} + 1$ M1

$2 {(\frac{d y}{d x})}^{2} + 2 y \frac{d^{2} y}{d x^{2}} = 6 x$ A1

recognizes that $\frac{d^{2} y}{d x^{2}} = 0$ (M1)

$\frac{d y}{d x} = \pm \sqrt{3 x}$

$(\pm) \frac{3 x^{2} + 1}{2 \sqrt{x^{3} + x}} = (\pm) \sqrt{3 x}$ (A1)

THEN

$12 x (x + x^{3}) = {(1 + 3 x^{2})}^{2}$

$12 x^{2} + 12 x^{4} = 9 x^{4} + 6 x^{2} + 1$

$3 x^{4} + 6 x^{2} - 1 = 0$ A1

attempt to use quadratic formula or equivalent (M1)

$x^{2} = \frac{- 6 \pm \sqrt{48}}{6}$

$(x > 0 \Rightarrow) x = \sqrt{\frac{2 \sqrt{3} - 3}{3}} (p = 2, q = - 3, r = 3)$ A1

Note: Accept any integer multiple of $p, q$ and $r$ (e.g. $4, - 6$ and $6$ ).

[7 marks]

attempt to find tangent line through $(- 1, - 1)$ (M1)

$y + 1 = - \frac{3}{2} (x + 1)$ OR $y = - 1.5 x - 2.5$ A1

[2 marks]

f.i.

attempt to solve simultaneously with $y^{2} = x^{3} + 2$ (M1)

Note: The M1 mark can be awarded for an unsupported correct answer in an incorrect format (e.g. $(4.25, - 8.875)$ ).

obtain $(\frac{17}{4}, - \frac{71}{8})$ A1

[2 marks]

f.ii.

attempt to find equation of $[QS]$ (M1)

$\frac{y - 1}{x + 1} = - \frac{79}{42} (= - 1.88095 \dots)$ (A1)

solve simultaneously with $y^{2} = x^{3} + 2$ (M1)

$x = 0.28798 \dots (= \frac{127}{441})$ A1

$y = - 1.4226 \dots (= \frac{13175}{9261})$ A1

$(0.228, - 1.42)$

attempt to find vector equation of $[QS]$ (M1)

$(\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} - 1 \\ 1 \end{matrix}) + λ (\begin{matrix} \frac{21}{4} \\ - \frac{79}{8} \end{matrix})$ (A1)

$x = - 1 + \frac{21}{4} λ$

$y = 1 - \frac{79}{8} λ$

attempt to solve ${(1 - \frac{79}{8} λ)}^{2} = {(- 1 + \frac{21}{4} λ)}^{3} + 2$ (M1)

$λ = 0.2453 \dots$

$x = 0.28798 \dots (= \frac{127}{441})$ A1

$y = - 1.4226 \dots (= \frac{13175}{9261})$ A1

$(0.228, - 1.42)$

[5 marks]

Examiners report

This was a relatively straightforward start, though it was disappointing to see so many candidates sketch their graphs on two separate axes, despite the question stating they should be sketched on the same axes.

a.i.

[N/A]

a.ii.

Of those candidates producing clear sketches, the vast majority were able to recognise the points of inflexion and write down their coordinates. A small number embarked on a mostly fruitless algebraic approach rather than use their graphs as intended. The distinguishing features between curves tended to focus on points of intersection with the axes, which was accepted. Only a small number offered ideas such as $y \to \infty$ on both curves. A number of (incorrect) suggestions were seen, stating that both curves tended towards a linear asymptote.

b.i.

[N/A]

b.ii.

A majority of candidates' suggestions related to the number of intersection points with the coordinate axes, while the idea of the x-axis acting as a line of symmetry was also often seen.

The required differentiation was straightforward for the majority of candidates.

d.i.

[N/A]

d.ii.

The majority employed the quotient rule here, often doing so successfully to find a correct expression for $\frac{d^{2} y}{d x^{2}}$ . Despite realising that $\frac{d^{2} y}{d x^{2}} = 0$ , the resulting algebra to find the required solution proved a step too far for most. A number of slips were seen in candidates' working, though better candidates were able to answer the question confidently.

Mistakes proved to be increasingly common by this stage of the paper. Various equations of lines were suggested, with the incorrect $y = 1.5 x + 2.5$ appearing more than once. Only the better candidates were able to tackle the final part of the question with any success; it was pleasing to see a number of clear algebraic (only) approaches, though this was not necessary to obtain full marks.

f.i.

[N/A]

f.ii.

Significant work on this question part was rarely seen, and it may have been the case that many candidates chose to spend their remaining time on the second question, especially if they felt they were making little progress with part f. Having said that, correct final answers were seen from better candidates, though these were few and far between.

22M.3.AHL.TZ1.2b

Question

This question asks you to explore cubic polynomials of the form $(x - r) (x^{2} - 2 a x + a^{2} + b^{2})$ for $x \in ℝ$ and corresponding cubic equations with one real root and two complex roots of the form $(z - r) (z^{2} - 2 a z + a^{2} + b^{2}) = 0$ for $z \in ℂ$ .

In parts (a), (b) and (c), let $r = 1, a = 4$ and $b = 1$ .

Consider the equation $(z - 1) (z^{2} - 8 z + 17) = 0$ for $z \in ℂ$ .

Consider the function $f (x) = (x - 1) (x^{2} - 8 x + 17)$ for $x \in ℝ$ .

Consider the function $g (x) = (x - r) (x^{2} - 2 a x + a^{2} + b^{2})$ for $x \in ℝ$ where $r, a \in ℝ$ and $b \in ℝ, b > 0$ .

The equation $(z - r) (z^{2} - 2 a z + a^{2} + b^{2}) = 0$ for $z \in ℂ$ has roots $r$ and $a \pm b i$ where $r, a \in ℝ$ and $b \in ℝ, b > 0$ .

On the Cartesian plane, the points $C_{1} (a, \sqrt{g' (a)})$ and $C_{2} (a, - \sqrt{g' (a)})$ represent the real and imaginary parts of the complex roots of the equation $(z - r) (z^{2} - 2 a z + a^{2} + b^{2}) = 0$ .

The following diagram shows a particular curve of the form $y = (x - r) (x^{2} - 2 a x + a^{2} + 16)$ and the tangent to the curve at the point $A (a, 80)$ . The curve and the tangent both intersect the $x$ -axis at the point $R (- 2, 0)$ . The points $C_{1}$ and $C_{2}$ are also shown.

Consider the curve $y = (x - r) (x^{2} - 2 a x + a^{2} + b^{2})$ for $a \neq r, b > 0$ . The points $A (a, g (a))$ and $R (r, 0)$ are as defined in part (d)(ii). The curve has a point of inflexion at point $P$ .

Consider the special case where $a = r$ and $b > 0$ .

Given that $1$ and $4 + i$ are roots of the equation, write down the third root.

[1]

a.i.

Verify that the mean of the two complex roots is $4$ .

[1]

a.ii.

Show that the line $y = x - 1$ is tangent to the curve $y = f (x)$ at the point $A (4, 3)$ .

[4]

Sketch the curve $y = f (x)$ and the tangent to the curve at point $A$ , clearly showing where the tangent crosses the $x$ -axis.

[2]

Show that $g' (x) = 2 (x - r) (x - a) + x^{2} - 2 a x + a^{2} + b^{2}$ .

[2]

d.i.

Hence, or otherwise, prove that the tangent to the curve $y = g (x)$ at the point $A (a, g (a))$ intersects the $x$ -axis at the point $R (r, 0)$ .

[6]

d.ii.

Deduce from part (d)(i) that the complex roots of the equation $(z - r) (z^{2} - 2 a z + a^{2} + b^{2}) = 0$ can be expressed as $a \pm i \sqrt{g' (a)}$ .

[1]

Use this diagram to determine the roots of the corresponding equation of the form $(z - r) (z^{2} - 2 a z + a^{2} + 16) = 0$ for $z \in ℂ$ .

[4]

f.i.

State the coordinates of $C_{2}$ .

[1]

f.ii.

Show that the $x$ -coordinate of $P$ is $\frac{1}{3} (2 a + r)$ .

You are not required to demonstrate a change in concavity.

[2]

g.i.

Hence describe numerically the horizontal position of point $P$ relative to the horizontal positions of the points $R$ and $A$ .

[1]

g.ii.

Sketch the curve $y = (x - r) (x^{2} - 2 a x + a^{2} + b^{2})$ for $a = r = 1$ and $b = 2$ .

[2]

h.i.

For $a = r$ and $b > 0$ , state in terms of $r$ , the coordinates of points $P$ and $A$ .

[1]

h.ii.

Markscheme

$4 - i$ A1

[1 mark]

a.i.

mean $= \frac{1}{2} (4 + i + 4 - i)$ A1

$= 4$ AG

[1 mark]

a.ii.

METHOD 1

attempts product rule differentiation (M1)

Note: Award (M1) for attempting to express $f (x)$ as $f (x) = x^{3} - 9 x^{2} + 25 x - 17$

$f' (x) = (x - 1) (2 x - 8) + x^{2} - 8 x + 17 (f' (x) = 3 x^{2} - 18 x + 25)$ A1

$f' (4) = 1$ A1

Note: Where $f' (x)$ is correct, award A1 for solving $f' (x) = 1$ and obtaining $x = 4$ .

EITHER

$y - 3 = 1 (x - 4)$ A1

$y = x + c$

$3 = 4 + c \Rightarrow c = - 1$ A1

states the gradient of $y = x - 1$ is also $1$ and verifies that $(4, 3)$ lies on the line $y = x - 1$ A1

THEN

so $y = x - 1$ is the tangent to the curve at $A (4, 3)$ AG

Note: Award a maximum of (M0)A0A1A1 to a candidate who does not attempt to find $f' (x)$ .

METHOD 2

sets $f (x) = x - 1$ to form $x - 1 = (x - 1) (x^{2} - 8 x + 17)$ (M1)

EITHER

$(x - 1) (x^{2} - 8 x + 16) = 0 (x^{3} - 9 x^{2} + 24 x - 16 = 0)$ A1

attempts to solve a correct cubic equation (M1)

$(x - 1) {(x - 4)}^{2} = 0 \Rightarrow x = 1, 4$

recognises that $x \neq 1$ and forms $x^{2} - 8 x + 17 = 1 (x^{2} - 8 x + 16 = 0)$ A1

attempts to solve a correct quadratic equation (M1)

${(x - 4)}^{2} = 0 \Rightarrow x = 4$

THEN

$x = 4$ is a double root R1

so $y = x - 1$ is the tangent to the curve at $A (4, 3)$ AG

Note: Candidates using this method are not required to verify that $y = 3$ .

[4 marks]

a positive cubic with an $x$ -intercept $(x = 1)$ , and a local maximum and local minimum in the first quadrant both positioned to the left of $A$ A1

Note: As the local minimum and point A are very close to each other, condone graphs that seem to show these points coinciding.
For the point of tangency, accept labels such as $A, (4, 3)$ or the point labelled from both axes. Coordinates are not required.

a correct sketch of the tangent passing through $A$ and crossing the $x$ -axis at the same point $(x = 1)$ as the curve A1

Note: Award A1A0 if both graphs cross the $x$ -axis at distinctly different points.

[2 marks]

EITHER

$g' (x) = (x - r) (2 x - 2 a) + x^{2} - 2 a x + a^{2} + b^{2}$ (M1)A1

$g (x) = x^{3} - (2 a + r) x^{2} + (a^{2} + b^{2} + 2 a r) x - (a^{2} + b^{2}) r$

attempts to find $g' (x)$ M1

$g' (x) = 3 x^{2} - 2 (2 a + r) x + a^{2} + b^{2} + 2 a r$

$= 2 x^{2} - 2 (a + r) x + 2 a r + x^{2} - 2 a x + a^{2} + b^{2}$ A1

$(= 2 (x^{2} - a x - r x + a r) + x^{2} - 2 a x + a^{2} + b^{2})$

THEN

$g' (x) = 2 (x - r) (x - a) + x^{2} - 2 a x + a^{2} + b^{2}$ AG

[2 marks]

d.i.

METHOD 1

$g (a) = b^{2} (a - r)$ (A1)

$g' (a) = b^{2}$ (A1)

attempts to substitute their $g (a)$ and $g' (a)$ into $y - g (a) = g' (a) (x - a)$ M1

$y - b^{2} (a - r) = b^{2} (x - a)$

EITHER

$y = b^{2} (x - r) (y = b^{2} x - b^{2} r)$ A1

sets $y = 0$ so $b^{2} (x - r) = 0$ M1

$b > 0 \Rightarrow x = r$ OR $b \neq 0 \Rightarrow x = r$ R1

sets $y = 0$ so $- b^{2} (a - r) = b^{2} (x - a)$ M1

$b > 0$ OR $b \neq 0 \Rightarrow - (a - r) = x - a$ R1

$x = r$ A1

THEN

so the tangent intersects the $x$ -axis at the point $R (r, 0)$ AG

METHOD 2

$g' (a) = b^{2}$ (A1)

$g (a) = b^{2} (a - r)$ (A1)

attempts to substitute their $g (a)$ and $g' (a)$ into $y = g' (a) x + c$ and attempts to find $c$ M1

$c = - b^{2} r$

EITHER

$y = b^{2} (x - r) (y = b^{2} x - b^{2} r)$ A1

sets $y = 0$ so $b^{2} (x - r) = 0$ M1

$b > 0 \Rightarrow x = r$ OR $b \neq 0 \Rightarrow x = r$ R1

sets $y = 0$ so $b^{2} (x - r) = 0$ M1

$b > 0$ OR $b \neq 0 \Rightarrow x - r = 0$ R1

$x = r$ A1

METHOD 3

$g' (a) = b^{2}$ (A1)

the line through $R (r, 0)$ parallel to the tangent at $A$ has equation
$y = b^{2} (x - r)$ A1

sets $g (x) = b^{2} (x - r)$ to form $b^{2} (x - r) = (x - r) (x^{2} - 2 a x + a^{2} + b^{2})$ M1

$b^{2} = x^{2} - 2 a x + a^{2} + b^{2}, (x \neq r)$ A1

${(x - a)}^{2} = 0$ A1

since there is a double root $(x = a)$ , this parallel line through $R (r, 0)$ is the required tangent at $A$ R1

[6 marks]

d.ii.

EITHER

$g' (a) = b^{2} \Rightarrow b = \sqrt{g' (a)}$ (since $b > 0$ ) R1

Note: Accept $b = \pm \sqrt{g' (a)}$ .

$(a \pm b i=) a \pm i \sqrt{b^{2}}$ and $g' (a) = b^{2}$ R1

THEN

hence the complex roots can be expressed as $a \pm i \sqrt{g' (a)}$ AG

[1 mark]

$b = 4$ (seen anywhere) A1

EITHER

attempts to find the gradient of the tangent in terms of $a$ and equates to $16$ (M1)

OR

substitutes $r = - 2, x = a$ and $y = 80$ to form $80 = (a - (- 2)) (a^{2} - 2 a^{2} + a^{2} + 16)$ (M1)

substitutes $r = - 2, x = a$ and $y = 80$ into $y = 16 (x - r)$ (M1)

THEN

$\frac{80}{a + 2} = 16 \Rightarrow a = 3$

roots are $- 2$ (seen anywhere) and $3 \pm 4 i$ A1A1

Note: Award A1 for $- 2$ and A1 for $3 \pm 4 i$ . Do not accept coordinates.

[4 marks]

f.i.

$(3, - 4)$ A1

Note: Accept “ $x = 3$ and $y = - 4$ ”.
Do not award A1FT for $(a, - 4)$ .

[1 mark]

f.ii.

$g' (x) = 2 (x - r) (x - a) + x^{2} - 2 a x + a^{2} + b^{2}$

attempts to find $g'' (x)$ M1

$g'' (x) = 2 (x - a) + 2 (x - r) + 2 x - 2 a (= 6 x - 2 r - 4 a)$

sets $g'' (x) = 0$ and correctly solves for $x$ A1

for example, obtaining $x - r + 2 (x - a) = 0$ leading to $3 x = 2 a + r$

so $x = \frac{1}{3} (2 a + r)$ AG

Note: Do not award A1 if the answer does not lead to the AG.

[2 marks]

g.i.

point $P$ is $\frac{2}{3}$ of the horizontal distance (way) from point $R$ to point $A$ A1

Note: Accept equivalent numerical statements or a clearly labelled diagram displaying the numerical relationship.
Award A0 for non-numerical statements such as “ $P$ is between $R$ and $A$ , closer to $A$ ”.

[1 mark]

g.ii.

$y = (x - 1) (x^{2} - 2 x + 5)$ (A1)

a positive cubic with no stationary points and a non-stationary point of inflexion at $x = 1$ A1

Note: Graphs may appear approximately linear. Award this A1 if a change of concavity either side of $x = 1$ is apparent.
Coordinates are not required and the $y$ -intercept need not be indicated.

[2 marks]

h.i.

$(r, 0)$ A1

[1 mark]

h.ii.

Examiners report

Part (a) (i) was generally well done with a significant majority of candidates using the conjugate root theorem to state $4 - i$ as the third root. A number of candidates, however, wasted considerable time attempting an algebraic method to determine the third root. Part (a) (ii) was reasonably well done. A few candidates however attempted to calculate the product of $4 + i$ and $4 - i$ .

Part (b) was reasonably well done by a significant number of candidates. Most were able to find a correct expression for $f' (x)$ and a good number of those candidates were able to determine that $f' (4) = 1$ . Candidates that did not determine the equation of the tangent had to state that the gradient of $y = x - 1$ is also 1 and verify that the point (4,3) lies on the line. A few candidates only met one of those requirements. Weaker candidates tended to only verify that the point (4,3) lies on the curve and the tangent line without attempting to find $f' (x)$ .

Part (c) was not answered as well as anticipated. A number of sketches were inaccurate and carelessly drawn with many showing both graphs crossing the x-axis at distinctly different points.

Part (d) (i) was reasonably well done by a good number of candidates. Most successful responses involved use of the product rule. A few candidates obtained full marks by firstly expanding $g (x)$ , then differentiating to find $g' (x)$ and finally simplifying to obtain the desired result. A number of candidates made elementary mistakes when differentiating. In general, the better candidates offered reasonable attempts at showing the general result in part (d) (ii). A good number gained partial credit by determining that $g' (a) = b^{2}$ and/or $g (a) = b^{2} (a - r)$ . Only the very best candidates obtained full marks by concluding that as $b > 0$ or $b \neq 0$ , then $x = r$ when $y = 0$ .

In general, only the best candidates were able to use the result $g' (a) = b^{2}$ to deduce that the complex roots of the equation can be expressed as $a \pm i \sqrt{g' (a)}$ . Although given the complex roots $a \pm b i$ , a significant number of candidates attempted, with mixed success, to use the quadratic formula to solve the equation $z^{2} - 2 a z + a^{2} + b^{2} = 0$ .

In part (f) (i), only a small number of candidates were able to determine all the roots of the equation. Disappointingly, a large number did not state $- 2$ as a root. Some candidates determined that $b = 4$ but were unable to use the diagram to determine that $a = 3$ . Of the candidates who determined all the roots in part (f) (i), very few gave the correct coordinates for C₂ . The most frequent error was to give the y-coordinate as $3 - 4 i$ .

Of the candidates who attempted part (g) (i), most were able to find an expression for $g'' (x)$ and a reasonable number of these were then able to convincingly show that $x = \frac{1}{3} (2 a + r)$ . It was very rare to see a correct response to part (g) (ii). A few candidates stated that P is between R and A with some stating that P was closer to A. A small number restated $x = \frac{1}{3} (2 a + r)$ in words.

Of the candidates who attempted part (h) (i), most were able to determine that $y = (x - 1) (x^{2} - 2 x + 5)$ . However, most graphs were poorly drawn with many showing a change in concavity at $x = 0$ rather than at $x = 1$ . In part (h) (ii), only a very small number of candidates determined that A and P coincide at (r,0).

a.i.

[N/A]

a.ii.

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

f.i.

[N/A]

f.ii.

[N/A]

g.i.

[N/A]

g.ii.

[N/A]

h.i.

[N/A]

h.ii.

EXM.3.AHL.TZ0.1a

Question

This question will investigate power series, as an extension to the Binomial Theorem for negative and fractional indices.

A power series in $x$ is defined as a function of the form $f (x) = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + . . .$ where the $a_{i} \in R$ .

It can be considered as an infinite polynomial.

This is an example of a power series, but is only a finite power series, since only a finite number of the $a_{i}$ are non-zero.

We will now attempt to generalise further.

Suppose ${(1 + x)}^{q}, q \in Q$ can be written as the power series $a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + . . .$ .

Expand ${(1 + x)}^{5}$ using the Binomial Theorem.

[2]

Consider the power series $1 - x + x^{2} - x^{3} + x^{4} - . . .$

By considering the ratio of consecutive terms, explain why this series is equal to ${(1 + x)}^{- 1}$ and state the values of $x$ for which this equality is true.

[4]

Differentiate the equation obtained part (b) and hence, find the first four terms in a power series for ${(1 + x)}^{- 2}$ .

[2]

Repeat this process to find the first four terms in a power series for ${(1 + x)}^{- 3}$ .

[2]

Hence, by recognising the pattern, deduce the first four terms in a power series for ${(1 + x)}^{- n}$ , $n \in Z^{+}$ .

[3]

By substituting $x = 0$ , find the value of $a_{0}$ .

[1]

By differentiating both sides of the expression and then substituting $x = 0$ , find the value of $a_{1}$ .

[2]

Repeat this procedure to find $a_{2}$ and $a_{3}$ .

[4]

Hence, write down the first four terms in what is called the Extended Binomial Theorem for ${(1 + x)}^{q}, q \in Q$ .

[1]

Write down the power series for $\frac{1}{1 + x^{2}}$ .

[2]

Hence, using integration, find the power series for $arctan x$ , giving the first four non-zero terms.

[4]

Markscheme

$1 + 5 x + 10 x^{2} + 10 x^{3} + 5 x^{4} + x^{5}$ M1A1

[2 marks]

It is an infinite GP with $a = 1, r = - x$ R1A1

$S_{\infty} = \frac{1}{1 - (- x)} = \frac{1}{1 + x} = {(1 + x)}^{- 1}$ M1A1AG

[4 marks]

${(1 + x)}^{- 1} = 1 - x + x^{2} - x^{3} + x^{4} - . . .$

$- 1 {(1 + x)}^{- 2} = - 1 + 2 x - 3 x^{2} + 4 x^{3} - . . .$ A1

${(1 + x)}^{- 2} = 1 - 2 x + 3 x^{2} - 4 x^{3} + . . .$ A1

[2 marks]

$- 2 {(1 + x)}^{- 3} = - 2 + 6 x - 12 x^{2} + 20 x^{3} . . .$ A1

${(1 + x)}^{- 3} = 1 - 3 x + 6 x^{2} - 10 x^{3} . . .$ A1

[2 marks]

${(1 + x)}^{- n} = 1 - n x + \frac{n (n + 1)}{2!} x^{2} - \frac{n (n + 1) (n + 2)}{3!} x^{3} . . .$ A1A1A1

[3 marks]

$1^{q} = a_{0} \Rightarrow a_{0} = 1$ A1

[1 mark]

$q {(1 + x)}^{q - 1} = a_{1} + 2 a_{2} x + 3 a_{3} x^{2} + . . .$ A1

$a_{1} = q$ A1

[2 marks]

$q (q - 1) {(1 + x)}^{q - 2} = 1 \times 2 a_{2} + 2 \times 3 a_{3} x + . . .$ A1

$a_{2} = \frac{q (q - 1)}{2!}$ A1

$q (q - 1) (q - 2) {(1 + x)}^{q - 3} = 1 \times 2 \times 3 a_{3} + . . .$ A1

$a_{3} = \frac{q (q - 1) (q - 2)}{3!}$ A1

[4 marks]

${(1 + x)}^{q} = 1 + q x + \frac{q (q - 1)}{2!} x^{2} + \frac{q (q - 1) (q - 2)}{3!} x^{3} . . .$ A1

[1 mark]

$\frac{1}{1 + x^{2}} = 1 - x^{2} + x^{4} - x^{6} + . . .$ M1A1

[2 marks]

$arctan x + c = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \frac{x^{7}}{7} + . . .$ M1A1

Putting $x = 0 \Rightarrow c = 0$ R1

So $arctan x = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \frac{x^{7}}{7} + . . .$ A1

[4 marks]

Examiners report

[N/A]

EXM.3.AHL.TZ0.3a

Question

This question will investigate methods for finding definite integrals of powers of trigonometrical functions.

Let $I_{n} = \int_{0}^{\frac{π}{2}} si n^{n} x d x, n \in N$ .

Let $J_{n} = \int_{0}^{\frac{π}{2}} co s^{n} x d x, n \in N .$

Let $T_{n} = \int_{0}^{\frac{π}{4}} ta n^{n} x d x, n \in N$ .

Find the exact values of $I_{0}$ , $I_{1}$ and $I_{2}$ .

[6]

Use integration by parts to show that $I_{n} = \frac{n - 1}{n} I_{n - 2}, n ⩾ 2$ .

[5]

b.i.

Explain where the condition $n ⩾ 2$ was used in your proof.

[1]

b.ii.

Hence, find the exact values of $I_{3}$ and $I_{4}$ .

[2]

Use the substitution $x = \frac{π}{2} - u$ to show that $J_{n} = I_{n}$ .

[4]

Hence, find the exact values of $J_{5}$ and $J_{6}$

[2]

Find the exact values of $T_{0}$ and $T_{1}$ .

[3]

Use the fact that $ta n^{2} x = se c^{2} x - 1$ to show that $T_{n} = \frac{1}{n - 1} - T_{n - 2}, n ⩾ 2$ .

[3]

g.i.

Explain where the condition $n ⩾ 2$ was used in your proof.

[1]

g.ii.

Hence, find the exact values of $T_{2}$ and $T_{3}$ .

[2]

Markscheme

$I_{0} = \int_{0}^{\frac{π}{2}} 1 d x = {[x]}_{0}^{\frac{π}{2}} = \frac{π}{2}$ M1A1

$I_{1} = \int_{0}^{\frac{π}{2}} sin x d x = {[- cos x]}_{0}^{\frac{π}{2}} = 1$ M1A1

$I_{2} = \int_{0}^{\frac{π}{2}} si n^{2} x d x = \int_{0}^{\frac{π}{2}} \frac{1 - cos 2 x}{2} d x = {[\frac{x}{2} - \frac{sin 2 x}{4}]}_{0}^{\frac{π}{2}} = \frac{π}{4}$ M1A1

[6 marks]

$u = si n^{n - 1} x$ $v = - \cos x$

$\frac{d u}{d x} = (n - 1) si n^{n - 2} x cos x$ $\frac{d v}{d x} = sin x$

$I_{n} = {[- si n^{n - 1} x cos x]}_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} (n - 1) si n^{n - 2} x cos^{2} x d x$ M1A1A1

$= 0 + \int_{0}^{\frac{π}{2}} (n - 1) si n^{n - 2} x (1 - si n^{2} x) d x = (n - 1) (I_{n - 2} - I_{n})$ M1A1

$\Rightarrow n I_{n} = (n - 1) I_{n - 2} \Rightarrow I_{n} = \frac{(n - 1)}{n} I_{n - 2}$ AG

[6 marks]

b.i.

need $n ⩾ 2$ so that $si n^{n - 1} \frac{π}{2} = 0$ in ${[- si n^{n - 1} x cos x]}_{0}^{\frac{π}{2}}$ R1

[1 mark]

b.ii.

$I_{3} = \frac{2}{3} I_{1} = \frac{2}{3} I_{4} = \frac{3}{4} I_{2} = \frac{3 π}{16}$ A1A1

[2 marks]

$x = \frac{π}{2} - u \Rightarrow \frac{d x}{d u} = - 1$ A1

$J_{n} = \int_{0}^{\frac{π}{2}} co s^{n} x d x = \int_{\frac{π}{2}}^{0} - co s^{n} (\frac{π}{2} - u) d u = - \int_{\frac{π}{2}}^{0} si n^{n} u d u = \int_{0}^{\frac{π}{2}} si n^{n} u d u = I_{n}$ M1A1A1AG

[4 marks]

$J_{5} = I_{5} = \frac{4}{5} I_{3} = \frac{4}{5} \times \frac{2}{3} = \frac{8}{15} J_{6} = I_{6} = \frac{5}{6} I_{4} = \frac{5}{6} \times \frac{3 π}{16} = \frac{5 π}{32}$ A1A1

[2 marks]

$T_{0} = \int_{0}^{\frac{π}{4}} 1 d x = {[x]}_{0}^{\frac{π}{4}} = \frac{π}{4}$ A1

$T_{1} = \int_{0}^{\frac{π}{4}} tan d x = {[- \ln | cos x |]}_{0}^{\frac{π}{4}} = - ln \frac{1}{\sqrt{2}} = ln \sqrt{2}$ M1A1

[3 marks]

$T_{n} = \int_{0}^{\frac{π}{4}} ta n^{n} x d x = \int_{0}^{\frac{π}{4}} ta n^{n - 2} x ta n^{2} x d x = \int_{0}^{\frac{π}{4}} ta n^{n - 2} x (se c^{2} x - 1) d x$ M1

$\int_{0}^{\frac{π}{4}} ta n^{n - 2} x se c^{2} x d x - \int_{0}^{\frac{π}{4}} ta n^{n - 2} x d x = {[\frac{ta n^{n - 1} x}{n - 1}]}_{0}^{\frac{π}{4}} - T_{n - 2} = \frac{1}{n - 1} - T_{n - 2}$ A1A1AG

[3 marks]

g.i.

need $n ⩾ 2$ so that the powers of tan in $\int_{0}^{\frac{π}{4}} ta n^{n - 2} x se c^{2} x d x - \int_{0}^{\frac{π}{4}} ta n^{n - 2} x d x$ are not negative R1

[1 mark]

g.ii.

$T_{2} = 1 - T_{0} = 1 - \frac{π}{4}$ A1

$T_{3} = \frac{1}{2} - T_{1} = \frac{1}{2} - \ln \sqrt{2}$ A1

[2 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

g.i.

[N/A]

g.ii.

[N/A]

EXM.3.AHL.TZ0.4a

Question

This question investigates some applications of differential equations to modeling population growth.

One model for population growth is to assume that the rate of change of the population is proportional to the population, i.e. $\frac{d P}{d t} = k P$ , where $k \in R$ , $t$ is the time (in years) and $P$ is the population

The initial population is 1000.

Given that $k = 0.003$ , use your answer from part (a) to find

Consider now the situation when $k$ is not a constant, but a function of time.

Given that $k = 0.003 + 0.002 t$ , find

Another model for population growth assumes

there is a maximum value for the population, $L$ .
that $k$ is not a constant, but is proportional to $(1 - \frac{P}{L})$ .

Show that the general solution of this differential equation is $P = A e^{k t}$ , where $A \in R$ .

[5]

the population after 10 years

[2]

b.i.

the number of years it will take for the population to triple.

[2]

b.ii.

$lim_{t \to \infty} P$

[1]

b.iii.

the solution of the differential equation, giving your answer in the form $P = f (t)$ .

[5]

c.i.

the number of years it will take for the population to triple.

[4]

c.ii.

Show that $\frac{d P}{d t} = \frac{m}{L} P (L - P)$ , where $m \in R$ .

[2]

Solve the differential equation $\frac{d P}{d t} = \frac{m}{L} P (L - P)$ , giving your answer in the form $P = g (t)$ .

[10]

Given that the initial population is 1000, $L = 10000$ and $m = 0.003$ , find the number of years it will take for the population to triple.

[4]

Markscheme

$\int \frac{1}{P} d P = \int k d t$ M1A1

$ln P = k t + c$ A1A1

$P = e^{k t + c}$ A1

$P = A e^{k t}$ , where $A = e^{c}$ AG

[5 marks]

when $t = 0, P = 1000$

$\Rightarrow A = 1000$ A1

$P (10) = 1000 e^{0.003 (10)} = 1030$ A1

[2 marks]

b.i.

$3000 = 1000 e^{0.003 t}$ M1

$t = \frac{ln 3}{0.003} = 366$ years A1

[2 marks]

b.ii.

$lim_{t \to \infty} P = \infty$ A1

[1 mark]

b.iii.

$\int \frac{1}{P} d P = \int (0.003 + 0.002 t) d t$ M1

$ln P = 0.003 t + 0.001 t^{2} + c$ A1A1

$P = e^{0.003 t + 0.001 t^{2} + c}$ A1

when $t = 0, P = 1000$

$\Rightarrow e^{c} = 1000$ M1

$P = 1000 e^{0.003 t + 0.001 t^{2}}$

[5 marks]

c.i.

$3000 = 1000 e^{0.003 t + 0.001 t^{2}}$ M1

$ln 3 = 0.003 t + 0.001 t^{2}$ A1

Use of quadratic formula or GDC graph or GDC polysmlt M1

$t = 31.7$ years A1

[4 marks]

c.ii.

$k = m (1 - \frac{P}{L})$ , where $m$ is the constant of proportionality A1

So $\frac{d P}{d t} = m (1 - \frac{P}{L}) P$ A1

$\frac{d P}{d t} = \frac{m}{L} P (L - P)$ AG

[2 marks]

$\int \frac{1}{P (L - P)} d P = \int \frac{m}{L} d t$ M1

$\frac{1}{P (L - P)} = \frac{A}{P} + \frac{B}{L - P}$ M1

$1 \equiv A (L - P) + B P$ A1

$A = \frac{1}{L}, B = \frac{1}{L}$ A1

$\frac{1}{L} \int (\frac{1}{P} + \frac{1}{L - P}) d P = \int \frac{m}{L} d t$

$\frac{1}{L} (ln P - ln (L - P)) = \frac{m}{L} t + c$ A1A1

$ln (\frac{P}{L - P}) = m t + d$ , where $d = c L$ M1

$\frac{P}{L - P} = C e^{m t}$ , where $C = e^{d}$ A1

$P (1 + C e^{m t}) = C L e^{m t}$ M1

$P = \frac{C L e^{m t}}{(1 + C e^{m t})} (= \frac{L}{(D e^{- m t} + 1)}, where D = \frac{1}{C})$ A1

[10 marks]

$1000 = \frac{10000}{D + 1}$ M1

$D = 9$ A1

$3000 = \frac{10000}{9 e^{- 0.003 t} + 1}$ M1

$t = 450$ years A1

[4 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

SPM.3.AHL.TZ0.1a

Question

This question asks you to investigate regular $n$ -sided polygons inscribed and circumscribed in a circle, and the perimeter of these as $n$ tends to infinity, to make an approximation for $π$ .

Let $P_{i} (n)$ represent the perimeter of any $n$ -sided regular polygon inscribed in a circle of radius 1 unit.

Consider an equilateral triangle ABC of side length, $x$ units, circumscribed about a circle of radius 1 unit and centre O as shown in the following diagram.

Let $P_{c} (n)$ represent the perimeter of any $n$ -sided regular polygon circumscribed about a circle of radius 1 unit.

Consider an equilateral triangle ABC of side length, $x$ units, inscribed in a circle of radius 1 unit and centre O as shown in the following diagram.

The equilateral triangle ABC can be divided into three smaller isosceles triangles, each subtending an angle of $\frac{2 π}{3}$ at O, as shown in the following diagram.

Using right-angled trigonometry or otherwise, show that the perimeter of the equilateral triangle ABC is equal to $3 \sqrt{3}$ units.

[3]

Consider a square of side length, $x$ units, inscribed in a circle of radius 1 unit. By dividing the inscribed square into four isosceles triangles, find the exact perimeter of the inscribed square.

[3]

Find the perimeter of a regular hexagon, of side length, $x$ units, inscribed in a circle of radius 1 unit.

[2]

Show that $P_{i} (n) = 2 n sin (\frac{π}{n})$ .

[3]

Use an appropriate Maclaurin series expansion to find $lim_{n \to \infty} P_{i} (n)$ and interpret this result geometrically.

[5]

Show that $P_{c} (n) = 2 n tan (\frac{π}{n})$ .

[4]

By writing $P_{c} (n)$ in the form $\frac{2 tan (\frac{π}{n})}{\frac{1}{n}}$ , find $lim_{n \to \infty} P_{c} (n)$ .

[5]

Use the results from part (d) and part (f) to determine an inequality for the value of $π$ in terms of $n$ .

[2]

The inequality found in part (h) can be used to determine lower and upper bound approximations for the value of $π$ .

Determine the least value for $n$ such that the lower bound and upper bound approximations are both within 0.005 of $π$ .

[3]

Markscheme

METHOD 1

consider right-angled triangle OCX where CX $= \frac{x}{2}$

$sin \frac{π}{3} = \frac{\frac{x}{2}}{1}$ M1A1

$\Rightarrow \frac{x}{2} = \frac{\sqrt{3}}{2} \Rightarrow x = \sqrt{3}$ A1

$P_{i} = 3 \times x = 3 \sqrt{3}$ AG

METHOD 2

eg use of the cosine rule $x^{2} = 1^{2} + 1^{2} - 2 (1) (1) cos \frac{2 π}{3}$ M1A1

$x = \sqrt{3}$ A1

$P_{i} = 3 \times x = 3 \sqrt{3}$ AG

Note: Accept use of sine rule.

[3 marks]

$sin \frac{π}{4} = \frac{1}{x}$ where $x$ = side of square M1

$x = \sqrt{2}$ A1

$P_{i} = 4 \sqrt{2}$ A1

[3 marks]

6 equilateral triangles ⇒ $x$ = 1 A1

$P_{i} = 6$ A1

[2 marks]

in right-angled triangle $sin (\frac{π}{n}) = \frac{\frac{x}{2}}{1}$ M1

$\Rightarrow x = 2 sin (\frac{π}{n})$ A1

$P_{i} = n \times x$

$P_{i} = n \times 2 sin (\frac{π}{n})$ M1

$P_{i} = 2 n sin (\frac{π}{n})$ AG

[3 marks]

consider $lim_{n \to \infty} 2 n sin (\frac{π}{n})$

use of $sin x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \dots$ M1

$2 n sin (\frac{π}{n}) = 2 n (\frac{π}{n} - \frac{π^{3}}{6 n^{3}} + \frac{π^{5}}{120 n^{5}} - \dots)$ (A1)

$= 2 (π - \frac{π^{3}}{6 n^{2}} + \frac{π^{5}}{120 n^{4}} - \dots)$ A1

$\Rightarrow lim_{n \to \infty} 2 n sin (\frac{π}{n}) = 2 π$ A1

as $n \to \infty$ polygon becomes a circle of radius 1 and $P_{i} = 2 π$ R1

[5 marks]

consider an $n$ -sided polygon of side length $x$

2 $n$ right-angled triangles with angle $\frac{2 π}{2 n} = \frac{π}{n}$ at centre M1A1

opposite side $\frac{x}{2} = tan (\frac{π}{n}) \Rightarrow x = 2 tan (\frac{π}{n})$ M1A1

Perimeter $P_{c} = 2 n tan (\frac{π}{n})$ AG

[4 marks]

consider $lim_{n \to \infty} 2 n tan (\frac{π}{n}) = lim_{n \to \infty} (\frac{2 tan (\frac{π}{n})}{\frac{1}{n}})$

$= lim_{n \to \infty} (\frac{2 tan (\frac{π}{n})}{\frac{1}{n}}) = \frac{0}{0}$ R1

attempt to use L’Hopital’s rule M1

$= lim_{n \to \infty} (\frac{- \frac{2 π}{n^{2}} se c^{2} (\frac{π}{n})}{- \frac{1}{n^{2}}})$ A1A1

$= 2 π$ A1

[5 marks]

$P_{i} < 2 π < P_{c}$

$2 n sin (\frac{π}{n}) < 2 π < 2 n tan (\frac{π}{n})$ M1

$n sin (\frac{π}{n}) < π < n tan (\frac{π}{n})$ A1

[2 marks]

attempt to find the lower bound and upper bound approximations within 0.005 of $π$ (M1)

$n$ = 46 A2

[3 marks]

Examiners report

[N/A]

SPM.3.AHL.TZ0.2a

Question

This question asks you to investigate some properties of the sequence of functions of the form $f_{n} (x) = cos (n arccos x)$ , −1 ≤ $x$ ≤ 1 and $n \in Z^{+}$ .

Important: When sketching graphs in this question, you are not required to find the coordinates of any axes intercepts or the coordinates of any stationary points unless requested.

For odd values of $n$ > 2, use your graphic display calculator to systematically vary the value of $n$ . Hence suggest an expression for odd values of $n$ describing, in terms of $n$ , the number of

For even values of $n$ > 2, use your graphic display calculator to systematically vary the value of $n$ . Hence suggest an expression for even values of $n$ describing, in terms of $n$ , the number of

The sequence of functions, $f_{n} (x)$ , defined above can be expressed as a sequence of polynomials of degree $n$ .

Consider $f_{n + 1} (x) = cos ((n + 1) arccos x)$ .

On the same set of axes, sketch the graphs of $y = f_{1} (x)$ and $y = f_{3} (x)$ for −1 ≤ $x$ ≤ 1.

[2]

local maximum points;

[3]

b.i.

local minimum points;

[1]

b.ii.

On a new set of axes, sketch the graphs of $y = f_{2} (x)$ and $y = f_{4} (x)$ for −1 ≤ $x$ ≤ 1.

[2]

local maximum points;

[3]

d.i.

local minimum points.

[1]

d.ii.

Solve the equation ${f_{n}}^{'} (x) = 0$ and hence show that the stationary points on the graph of $y = f_{n} (x)$ occur at $x = cos \frac{k π}{n}$ where $k \in Z^{+}$ and 0 < $k$ < $n$ .

[4]

Use an appropriate trigonometric identity to show that $f_{2} (x) = 2 x^{2} - 1$ .

[2]

Use an appropriate trigonometric identity to show that $f_{n + 1} (x) = cos (n arccos x) cos (arccos x) - sin (n arccos x) sin (arccos x)$ .

[2]

Hence show that $f_{n + 1} (x) + f_{n - 1} (x) = 2 x f_{n} (x)$ , $n \in Z^{+}$ .

[3]

h.i.

Hence express $f_{3} (x)$ as a cubic polynomial.

[2]

h.ii.

Markscheme

correct graph of $y = f_{1} (x)$ A1

correct graph of $y = f_{3} (x)$ A1

[2 marks]

graphical or tabular evidence that $n$ has been systematically varied M1

eg $n$ = 3, 1 local maximum point and 1 local minimum point

$n$ = 5, 2 local maximum points and 2 local minimum points

$n$ = 7, 3 local maximum points and 3 local minimum points (A1)

$\frac{n - 1}{2}$ local maximum points A1

[3 marks]

b.i.

$\frac{n - 1}{2}$ local minimum points A1

Note: Allow follow through from an incorrect local maximum formula expression.

[1 mark]

b.ii.

correct graph of $y = f_{2} (x)$ A1

correct graph of $y = f_{4} (x)$ A1

[2 marks]

graphical or tabular evidence that $n$ has been systematically varied M1

eg $n$ = 2, 0 local maximum point and 1 local minimum point

$n$ = 4, 1 local maximum points and 2 local minimum points

$n$ = 6, 2 local maximum points and 3 local minimum points (A1)

$\frac{n - 2}{2}$ local maximum points A1

[3 marks]

d.i.

$\frac{n}{2}$ local minimum points A1

[1 mark]

d.ii.

$f_{n} (x) = cos (n arccos (x))$

${f_{n}}^{'} (x) = \frac{n sin (n arccos (x))}{\sqrt{1 - x^{2}}}$ M1A1

Note: Award M1 for attempting to use the chain rule.

${f_{n}}^{'} (x) = 0 \Rightarrow n sin (n arccos (x)) = 0$ M1

$n arccos (x) = k π (k \in Z^{+})$ A1

leading to

$x = cos \frac{k π}{n}$ ( $k \in Z^{+}$ and 0 < $k$ < $n$ ) AG

[4 marks]

$f_{2} (x) = cos (2 arccos x)$

$= 2 {(cos (arccos x))}^{2} - 1$ M1

stating that $(cos (arccos x)) = x$ A1

so $f_{2} (x) = 2 x^{2} - 1$ AG

[2 marks]

$f_{n + 1} (x) = cos ((n + 1) arccos x)$

$= cos (n arccos x + arccos x)$ A1

use of cos(A + B) = cos A cos B − sin A sin B leading to M1

$= cos (n arccos x) cos (arccos x) - sin (n arccos x) sin (arccos x)$ AG

[2 marks]

$f_{n - 1} (x) = cos ((n - 1) arccos x)$ A1

$= cos (n arccos x) cos (arccos x) + sin (n arccos x) sin (arccos x)$ M1

$f_{n + 1} (x) + f_{n - 1} (x) = 2 cos (n arccos x) cos (arccos x)$ A1

$= 2 x f_{n} (x)$ AG

[3 marks]

h.i.

$f_{3} (x) = 2 x f_{2} (x) - f_{1} (x)$ (M1)

$= 2 x (2 x^{2} - 1) - x$

$= 4 x^{3} - 3 x$ A1

[2 marks]

h.ii.

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

h.i.

[N/A]

h.ii.

16N.1.SL.TZ0.S_6

Question

Let $f^{'} (x) = \sin^{3} (2 x) \cos (2 x)$ . Find $f (x)$ , given that $f (\frac{π}{4}) = 1$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

evidence of integration (M1)

eg $\int f^{'} (x) d x$

correct integration (accept missing $C$ ) (A2)

eg $\frac{1}{2} \times \frac{\sin^{4} (2 x)}{4}, \frac{1}{8} \sin^{4} (2 x) + C$

substituting initial condition into their integrated expression (must have $+ C$ ) M1

eg $1 = \frac{1}{8} \sin^{4} (\frac{π}{2}) + C$

Note: Award M0 if they substitute into the original or differentiated function.

recognizing $\sin (\frac{π}{2}) = 1$ (A1)

eg $1 = \frac{1}{8} (1)^{4} + C$

$C = \frac{7}{8}$ (A1)

$f (x) = \frac{1}{8} \sin^{4} (2 x) + \frac{7}{8}$ A1 N5

[7 marks]

Examiners report

[N/A]

16N.1.SL.TZ0.S_10a

Question

Let $f (x) = \cos x$ .

Let $g (x) = x^{k}$ , where $k \in Z^{+}$ .

Let $k = 21$ and $h (x) = (f^{(19)} (x) \times g^{(19)} (x))$ .

(i) Find the first four derivatives of $f (x)$ .

(ii) Find $f^{(19)} (x)$ .

[4]

(i) Find the first three derivatives of $g (x)$ .

(ii) Given that $g^{(19)} (x) = \frac{k!}{(k - p)!} (x^{k - 19})$ , find $p$ .

[5]

(i) Find $h^{'} (x)$ .

(ii) Hence, show that $h^{'} (π) = \frac{- 21!}{2} π^{2}$ .

[7]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

(i) $f^{'} (x) = - \sin x, f^{″} (x) = - \cos x, f^{(3)} (x) = \sin x, f^{(4)} (x) = \cos x$ A2 N2

(ii) valid approach (M1)

egrecognizing that 19 is one less than a multiple of 4, $f^{(19)} (x) = f^{(3)} (x)$

$f^{(19)} (x) = \sin x$ A1 N2

[4 marks]

(i) $g^{'} (x) = k x^{k - 1}$

$g^{″} (x) = k (k - 1) x^{k - 2}, g^{(3)} (x) = k (k - 1) (k - 2) x^{k - 3}$ A1A1 N2

(ii) METHOD 1

correct working that leads to the correct answer, involving the correct expression for the 19th derivative A2

eg $k (k - 1) (k - 2) \dots (k - 18) \times \frac{(k - 19)!}{(k - 19)!},_{k} P_{19}$

$p = 19$ (accept $\frac{k!}{(k - 19)!} x^{k - 19}$ ) A1 N1

METHOD 2

correct working involving recognizing patterns in coefficients of first three derivatives (may be seen in part (b)(i)) leading to a general rule for 19th coefficient A2

eg $g^{″} = 2! (\begin{matrix} k \\ 2 \end{matrix}), k (k - 1) (k - 2) = \frac{k!}{(k - 3)!}, g^{(3)} (x) =_{k} P_{3} (x^{k - 3})$

$g^{(19)} (x) = 19! (\begin{matrix} k \\ 19 \end{matrix}), 19! \times \frac{k!}{(k - 19)! \times 19!},_{k} P_{19}$

$p = 19$ (accept $\frac{k!}{(k - 19)!} x^{k - 19}$ ) A1 N1

[5 marks]

(i) valid approach using product rule (M1)

eg $u v^{'} + v u^{'}, f^{(19)} g^{(20)} + f^{(20)} g^{(19)}$

correct 20th derivatives (must be seen in product rule) (A1)(A1)

eg $g^{(20)} (x) = \frac{21!}{(21 - 20)!} x, f^{(20)} (x) = \cos x$

$h^{'} (x) = \sin x (21! x) + \cos x (\frac{21!}{2} x^{2}) (accept \sin x (\frac{21!}{1!} x) + \cos x (\frac{21!}{2!} x^{2}))$ A1 N3

(ii) substituting $x = π$ (seen anywhere) (A1)

eg $f^{(19)} (π) g^{(20)} (π) + f^{(20)} (π) g^{(19)} (π), \sin π \frac{21!}{1!} π + \cos π \frac{21!}{2!} π^{2}$

evidence of one correct value for $\sin π$ or $\cos π$ (seen anywhere) (A1)

eg $\sin π = 0, \cos π = - 1$

evidence of correct values substituted into $h^{'} (π)$ A1

eg $21! (π) (0 - \frac{π}{2!}), 21! (π) (- \frac{π}{2}), 0 + (- 1) \frac{21!}{2} π^{2}$

Note: If candidates write only the first line followed by the answer, award A1A0A0.

$\frac{- 21!}{2} π^{2}$ AG N0

[7 marks]

Examiners report

[N/A]

16N.1.SL.TZ0.T_14a

Question

The equation of a curve is $y = \frac{1}{2} x^{4} - \frac{3}{2} x^{2} + 7$ .

The gradient of the tangent to the curve at a point P is $- 10$ .

Find $\frac{d y}{d x}$ .

[2]

Find the coordinates of P.

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$2 x^{3} - 3 x$ (A1)(A1) (C2)

Note: Award (A1) for $2 x^{3}$ , award (A1) for $- 3 x$ .

Award at most (A1)(A0) if there are any extra terms.

[2 marks]

$2 x^{3} - 3 x = - 10$ (M1)

Note: Award (M1) for equating their answer to part (a) to $- 10$ .

$x = - 2$ (A1)(ft)

Note: Follow through from part (a). Award (M0)(A0) for $- 2$ seen without working.

$y = \frac{1}{2} (- 2)^{4} - \frac{3}{2} (- 2)^{2} + 7$ (M1)

Note: Award (M1) substituting their $- 2$ into the original function.

$y = 9$ (A1)(ft) (C4)

Note: Accept $(- 2, 9)$ .

[4 marks]

Examiners report

[N/A]

17M.1.SL.TZ2.S_5

Question

Let $f^{'} (x) = \frac{3 x^{2}}{{(x^{3} + 1)}^{5}}$ . Given that $f (0) = 1$ , find $f (x)$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid approach (M1)

eg $\int f^{'} d x, \int \frac{3 x^{2}}{{(x^{3} + 1)}^{5}} d x$

correct integration by substitution/inspection A2

eg $f (x) = - \frac{1}{4} (x^{3} + 1)^{- 4} + c, \frac{- 1}{4 {(x^{3} + 1)}^{4}}$

correct substitution into their integrated function (must include $c$ ) M1

eg $1 = \frac{- 1}{4 {(0^{3} + 1)}^{4}} + c, - \frac{1}{4} + c = 1$

Note: Award M0 if candidates substitute into $f^{'}$ or $f^{″}$ .

$c = \frac{5}{4}$ (A1)

$f (x) = - \frac{1}{4} (x^{3} + 1)^{- 4} + \frac{5}{4} (= \frac{- 1}{4 {(x^{3} + 1)}^{4}} + \frac{5}{4}, \frac{5 {(x^{3} + 1)}^{4} - 1}{4 {(x^{3} + 1)}^{4}})$ A1 N4

[6 marks]

Examiners report

[N/A]

17M.1.SL.TZ1.S_5a

Question

Find $\int x e^{x^{2} - 1} d x$ .

[4]

Find $f (x)$ , given that $f^{'} (x) = x e^{x^{2} - 1}$ and $f (- 1) = 3$ .

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid approach to set up integration by substitution/inspection (M1)

eg $u = x^{2} - 1, d u = 2 x, \int 2 x e^{x^{2} - 1} d x$

correct expression (A1)

eg $\frac{1}{2} \int 2 x e^{x^{2} - 1} d x, \frac{1}{2} \int e^{u} d u$

$\frac{1}{2} e^{x^{2} - 1} + c$ A2 N4

Notes: Award A1 if missing “ $+ c$ ”.

[4 marks]

substituting $x = - 1$ into their answer from (a) (M1)

eg $\frac{1}{2} e^{0}, \frac{1}{2} e^{1 - 1} = 3$

correct working (A1)

eg $\frac{1}{2} + c = 3, c = 2.5$

$f (x) = \frac{1}{2} e^{x^{2} - 1} + 2.5$ A1 N2

[3 marks]

Examiners report

[N/A]

17M.1.SL.TZ1.S_9c

Question

A quadratic function $f$ can be written in the form $f (x) = a (x - p) (x - 3)$ . The graph of $f$ has axis of symmetry $x = 2.5$ and $y$ -intercept at $(0, - 6)$

Find the value of $p$ .

[3]

Find the value of $a$ .

[3]

The line $y = k x - 5$ is a tangent to the curve of $f$ . Find the values of $k$ .

[8]

Markscheme

METHOD 1 (using x-intercept)

determining that 3 is an $x$ -intercept (M1)

eg $x - 3 = 0$ , M17/5/MATME/SP1/ENG/TZ1/09.a/M

valid approach (M1)

eg $3 - 2.5, \frac{p + 3}{2} = 2.5$

$p = 2$ A1 N2

METHOD 2 (expanding f (x))

correct expansion (accept absence of $a$ ) (A1)

eg $a x^{2} - a (3 + p) x + 3 a p, x^{2} - (3 + p) x + 3 p$

valid approach involving equation of axis of symmetry (M1)

eg $\frac{- b}{2 a} = 2.5, \frac{a (3 + p)}{2 a} = \frac{5}{2}, \frac{3 + p}{2} = \frac{5}{2}$

$p = 2$ A1 N2

METHOD 3 (using derivative)

correct derivative (accept absence of $a$ ) (A1)

eg $a (2 x - 3 - p), 2 x - 3 - p$

valid approach (M1)

eg $f^{'} (2.5) = 0$

$p = 2$ A1 N2

[3 marks]

attempt to substitute $(0, - 6)$ (M1)

eg $- 6 = a (0 - 2) (0 - 3), a (0)^{2} - 5 a (0) + 6 a = - 6$

correct working (A1)

eg $- 6 = 6 a$

$a = - 1$ A1 N2

[3 marks]

METHOD 1 (using discriminant)

recognizing tangent intersects curve once (M1)

recognizing one solution when discriminant = 0 M1

attempt to set up equation (M1)

eg $g = f, k x - 5 = - x^{2} + 5 x - 6$

rearranging their equation to equal zero (M1)

eg $x^{2} - 5 x + k x + 1 = 0$

correct discriminant (if seen explicitly, not just in quadratic formula) A1

eg $(k - 5)^{2} - 4, 25 - 10 k + k^{2} - 4$

correct working (A1)

eg $k - 5 = \pm 2, (k - 3) (k - 7) = 0, \frac{10 \pm \sqrt{100 - 4 \times 21}}{2}$

$k = 3, 7$ A1A1 N0

METHOD 2 (using derivatives)

attempt to set up equation (M1)

eg $g = f, k x - 5 = - x^{2} + 5 x - 6$

recognizing derivative/slope are equal (M1)

eg $f^{'} = m_{T}, f^{'} = k$

correct derivative of $f$ (A1)

eg $- 2 x + 5$

attempt to set up equation in terms of either $x$ or $k$ M1

eg $(- 2 x + 5) x - 5 = - x^{2} + 5 x - 6, k (\frac{5 - k}{2}) - 5 = - {(\frac{5 - k}{2})}^{2} + 5 (\frac{5 - k}{2}) - 6$

rearranging their equation to equal zero (M1)

eg $x^{2} - 1 = 0, k^{2} - 10 k + 21 = 0$

correct working (A1)

eg $x = \pm 1, (k - 3) (k - 7) = 0, \frac{10 \pm \sqrt{100 - 4 \times 21}}{2}$

$k = 3, 7$ A1A1 N0

[8 marks]

Examiners report

[N/A]

17M.1.SL.TZ1.S_10a

Question

The following table shows the probability distribution of a discrete random variable $A$ , in terms of an angle $θ$ .

M17/5/MATME/SP1/ENG/TZ1/10

Show that $\cos θ = \frac{3}{4}$ .

[6]

Given that $\tan θ > 0$ , find $\tan θ$ .

[3]

Let $y = \frac{1}{\cos x}$ , for $0 < x < \frac{π}{2}$ . The graph of $y$ between $x = θ$ and $x = \frac{π}{4}$ is rotated 360° about the $x$ -axis. Find the volume of the solid formed.

[6]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

evidence of summing to 1 (M1)

eg $\sum p = 1$

correct equation A1

eg $\cos θ + 2 \cos 2 θ = 1$

correct equation in $\cos θ$ A1

eg $\cos θ + 2 (2 \cos^{2} θ - 1) = 1, 4 \cos^{2} θ + \cos θ - 3 = 0$

evidence of valid approach to solve quadratic (M1)

egfactorizing equation set equal to $0, \frac{- 1 \pm \sqrt{1 - 4 \times 4 \times (- 3)}}{8}$

correct working, clearly leading to required answer A1

eg $(4 \cos θ - 3) (\cos θ + 1), \frac{- 1 \pm 7}{8}$

correct reason for rejecting $\cos θ \neq - 1$ R1

eg $\cos θ$ is a probability (value must lie between 0 and 1), $\cos θ > 0$

Note: Award R0 for $\cos θ \neq - 1$ without a reason.

$\cos θ = \frac{3}{4}$ AG N0

valid approach (M1)

egsketch of right triangle with sides 3 and 4, $\sin^{2} x + \cos^{2} x = 1$

correct working

(A1)

egmissing side $= \sqrt{7}, \frac{\frac{\sqrt{7}}{4}}{\frac{3}{4}}$

$\tan θ = \frac{\sqrt{7}}{3}$ A1 N2

[3 marks]

attempt to substitute either limits or the function into formula involving $f^{2}$ (M1)

eg $π \int_{θ}^{\frac{π}{4}} f^{2}, \int {(\frac{1}{\cos x})}^{2}$

correct substitution of both limits and function (A1)

eg $π \int_{θ}^{\frac{π}{4}} {(\frac{1}{\cos x})}^{2} d x$

correct integration (A1)

eg $\tan x$

substituting their limits into their integrated function and subtracting (M1)

eg $\tan \frac{π}{4} - \tan θ$

Note: Award M0 if they substitute into original or differentiated function.

$\tan \frac{π}{4} = 1$ (A1)

eg $1 - \tan θ$

$V = π - \frac{π \sqrt{7}}{3}$ A1 N3

[6 marks]

Examiners report

[N/A]

17M.1.SL.TZ2.T_13a

Question

The diagram shows part of the graph of a function $y = f (x)$ . The graph passes through point $A (1, 3)$ .

M17/5/MATSD/SP1/ENG/TZ2/13

The tangent to the graph of $y = f (x)$ at A has equation $y = - 2 x + 5$ . Let $N$ be the normal to the graph of $y = f (x)$ at A.

Write down the value of $f (1)$ .

[1]

Find the equation of $N$ . Give your answer in the form $a x + b y + d = 0$ where $a$ , $b$ , $d \in Z$ .

[3]

Draw the line $N$ on the diagram above.

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

3 (A1) (C1)

Notes: Accept $y = 3$

[1 mark]

$3 = 0.5 (1) + c$ OR $y - 3 = 0.5 (x - 1)$ (A1)(A1)

Note: Award (A1) for correct gradient, (A1) for correct substitution of $A (1, 3)$ in the equation of line.

$x - 2 y + 5 = 0$ or any integer multiple (A1)(ft) (C3)

Note: Award (A1)(ft) for their equation correctly rearranged in the indicated form.

The candidate’s answer must be an equation for this mark.

[3 marks]

M17/5/MATSD/SP1/ENG/TZ2/13.c/M (M1)(A1)(ft) (C2)

Note: Award M1) for a straight line, with positive gradient, passing through $(1, 3)$ , (A1)(ft) for line (or extension of their line) passing approximately through 2.5 or their intercept with the $y$ -axis.

[2 marks]

Examiners report

[N/A]

17N.1.SL.TZ0.S_6

Question

Let $f (x) = 15 - x^{2}$ , for $x \in R$ . The following diagram shows part of the graph of $f$ and the rectangle OABC, where A is on the negative $x$ -axis, B is on the graph of $f$ , and C is on the $y$ -axis.

N17/5/MATME/SP1/ENG/TZ0/06

Find the $x$ -coordinate of A that gives the maximum area of OABC.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to find the area of OABC (M1)

eg $OA \times OC, x \times f (x), f (x) \times (- x)$

correct expression for area in one variable (A1)

eg $area = x (15 - x^{2}), 15 x - x^{3}, x^{3} - 15 x$

valid approach to find maximum area (seen anywhere) (M1)

eg $A^{'} (x) = 0$

correct derivative A1

eg $15 - 3 x^{2}, (15 - x^{2}) + x (- 2 x) = 0, - 15 + 3 x^{2}$

correct working (A1)

eg $15 = 3 x^{2}, x^{2} = 5, x = \sqrt{5}$

$x = - \sqrt{5} (acceptA (- \sqrt{5}, 0))$ A2 N3

[7 marks]

Examiners report

[N/A]

17N.1.SL.TZ0.S_7

Question

Consider $f (x) = \log k (6 x - 3 x^{2})$ , for $0 < x < 2$ , where $k > 0$ .

The equation $f (x) = 2$ has exactly one solution. Find the value of $k$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1 – using discriminant

correct equation without logs (A1)

eg $6 x - 3 x^{2} = k^{2}$

valid approach (M1)

eg $- 3 x^{2} + 6 x - k^{2} = 0, 3 x^{2} - 6 x + k^{2} = 0$

recognizing discriminant must be zero (seen anywhere) M1

eg $Δ = 0$

correct discriminant (A1)

eg $6^{2} - 4 (- 3) (- k^{2}), 36 - 12 k^{2} = 0$

correct working (A1)

eg $12 k^{2} = 36, k^{2} = 3$

$k = \sqrt{3}$ A2 N2

METHOD 2 – completing the square

correct equation without logs (A1)

eg $6 x - 3 x^{2} = k^{2}$

valid approach to complete the square (M1)

eg $3 (x^{2} - 2 x + 1) = - k^{2} + 3, x^{2} - 2 x + 1 - 1 + \frac{k^{2}}{3} = 0$

correct working (A1)

eg $3 (x - 1)^{2} = - k^{2} + 3, (x - 1)^{2} - 1 + \frac{k^{2}}{3} = 0$

recognizing conditions for one solution M1

eg $(x - 1)^{2} = 0, - 1 + \frac{k^{2}}{3} = 0$

correct working (A1)

eg $\frac{k^{2}}{3} = 1, k^{2} = 3$

$k = \sqrt{3}$ A2 N2

[7 marks]

Examiners report

[N/A]

17N.1.SL.TZ0.S_8d

Question

Let $f (x) = x^{2} - x$ , for $x \in R$ . The following diagram shows part of the graph of $f$ .

N17/5/MATME/SP1/ENG/TZ0/08

The graph of $f$ crosses the $x$ -axis at the origin and at the point $P (1, 0)$ .

The line $L$ intersects the graph of $f$ at another point Q, as shown in the following diagram.

N17/5/MATME/SP1/ENG/TZ0/08.c.d

Find the area of the region enclosed by the graph of $f$ and the line $L$ .

Markscheme

valid approach (M1)

eg $\int L - f, \int_{- 1}^{1} (1 - x^{2}) d x$ , splitting area into triangles and integrals

correct integration (A1)(A1)

eg ${[x - \frac{x^{3}}{3}]}_{- 1}^{1}, - \frac{x^{3}}{3} - \frac{x^{2}}{2} + \frac{x^{2}}{2} + x$

substituting their limits into their integrated function and subtracting (in any order) (M1)

eg $1 - \frac{1}{3} - (- 1 - \frac{- 1}{3})$

Note: Award M0 for substituting into original or differentiated function.

area $= \frac{4}{3}$ A2 N3

[6 marks]

Examiners report

[N/A]

17N.1.SL.TZ0.T_7a

Question

Rosewood College has 120 students. The students can join the sports club ( $S$ ) and the music club ( $M$ ).

For a student chosen at random from these 120, the probability that they joined both clubs is $\frac{1}{4}$ and the probability that they joined the music club is $\frac{1}{3}$ .

There are 20 students that did not join either club.

Complete the Venn diagram for these students.

N17/5/MATSD/SP1/ENG/TZ0/07.a

[2]

One of the students who joined the sports club is chosen at random. Find the probability that this student joined both clubs.

[2]

Determine whether the events $S$ and $M$ are independent.

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

N17/5/MATSD/SP1/ENG/TZ0/07.a/M (A1)(A1) (C2)

Note: Award (A1) for 30 in correct area, (A1) for 60 and 10 in the correct areas.

[2 marks]

$\frac{30}{90} (\frac{1}{3}, 0.333333 \dots, 33.3333 \dots %)$ (A1)(ft)(A1)(ft) (C2)

Note: Award (A1)(ft) for correct numerator of 30, (A1)(ft) for correct denominator of 90. Follow through from their Venn diagram.

[2 marks]

$P (S) \times P (M) = \frac{3}{4} \times \frac{1}{3} = \frac{1}{4}$ (R1)

Note: Award (R1) for multiplying their by $\frac{1}{3}$ .

therefore the events are independent $(asP (S \cap M) = \frac{1}{4})$ (A1)(ft) (C2)

Note: Award (R1)(A1)(ft) for an answer which is consistent with their Venn diagram.

Do not award (R0)(A1)(ft).

Do not award final (A1) if $P (S) \times P (M)$ is not calculated. Follow through from part (a).

[2 marks]

Examiners report

[N/A]

17N.1.SL.TZ0.T_11a

Question

A quadratic function $f$ is given by $f (x) = a x^{2} + b x + c$ . The points $(0, 5)$ and $(- 4, 5)$ lie on the graph of $y = f (x)$ .

The $y$ -coordinate of the minimum of the graph is 3.

Find the equation of the axis of symmetry of the graph of $y = f (x)$ .

[2]

Write down the value of $c$ .

[1]

Find the value of $a$ and of $b$ .

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$x = - 2$ (A1)(A1) (C2)

Note: Award (A1) for $x =$ (a constant) and (A1) for $- 2$ .

[2 marks]

$(c =) 5$ (A1) (C1)

[1 mark]

$- \frac{b}{2 a} = - 2$

$a (- 2)^{2} - 2 b + 5 = 3$ or equivalent

$a (- 4)^{2} - 4 b + 5 = 5$ or equivalent

$2 a (- 2) + b = 0$ or equivalent (M1)

Note: Award (M1) for two of the above equations.

$a = 0.5$ (A1)(ft)

$b = 2$ (A1)(ft) (C3)

Note: Award at most (M1)(A1)(ft)(A0) if the answers are reversed.

Follow through from parts (a) and (b).

[3 marks]

Examiners report

[N/A]

17N.1.SL.TZ0.T_15a

Question

Maria owns a cheese factory. The amount of cheese, in kilograms, Maria sells in one week, $Q$ , is given by

$Q = 882 - 45 p$ ,

where $p$ is the price of a kilogram of cheese in euros (EUR).

Maria earns $(p - 6.80) EUR$ for each kilogram of cheese sold.

To calculate her weekly profit $W$ , in EUR, Maria multiplies the amount of cheese she sells by the amount she earns per kilogram.

Write down how many kilograms of cheese Maria sells in one week if the price of a kilogram of cheese is 8 EUR.

[1]

Find how much Maria earns in one week, from selling cheese, if the price of a kilogram of cheese is 8 EUR.

[2]

Write down an expression for $W$ in terms of $p$ .

[1]

Find the price, $p$ , that will give Maria the highest weekly profit.

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

522 (kg) (A1) (C1)

[1 mark]

$522 (8 - 6.80)$ or equivalent (M1)

Note: Award (M1) for multiplying their answer to part (a) by $(8 - 6.80)$ .

626 (EUR) (626.40) (A1)(ft) (C2)

Note: Follow through from part (a).

[2 marks]

$(W =) (882 - 45 p) (p - 6.80)$ (A1)

$(W =) - 45 p^{2} + 1188 p - 5997.6$ (A1) (C1)

[1 mark]

sketch of $W$ with some indication of the maximum (M1)

$- 90 p + 1188 = 0$ (M1)

Note: Award (M1) for equating the correct derivative of their part (c) to zero.

$(p =) \frac{- 1188}{2 \times (- 45)}$ (M1)

Note: Award (M1) for correct substitution into the formula for axis of symmetry.

$(p =) 13.2 (EUR)$ (A1)(ft) (C2)

Note: Follow through from their part (c), if the value of $p$ is such that $6.80 < p < 19.6$ .

[2 marks]

Examiners report

[N/A]

18M.1.SL.TZ1.S_5a

Question

Let $f (x) = \frac{1}{\sqrt{2 x - 1}}$ , for $x > \frac{1}{2}$ .

Find $\int {(f (x))}^{2} d x$ .

[3]

Part of the graph of f is shown in the following diagram.

The shaded region R is enclosed by the graph of f, the x-axis, and the lines x = 1 and x = 9 . Find the volume of the solid formed when R is revolved 360° about the x-axis.

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

correct working (A1)

eg $\int \frac{1}{2 x - 1} d x, \int {(2 x - 1)}^{- 1}, \frac{1}{2 x - 1}, \int {(\frac{1}{\sqrt{u}})}^{2} \frac{d u}{2}$

$\int {(f (x))}^{2} d x = \frac{1}{2} ln (2 x - 1) + c$ A2 N3

Note: Award A1 for $\frac{1}{2} ln (2 x - 1)$ .

[3 marks]

attempt to substitute either limits or the function into formula involving f ² (accept absence of $π$ / dx) (M1)

eg $\int_{1}^{9} y^{2} d x, π \int {(\frac{1}{\sqrt{2 x - 1}})}^{2} d x, {[\frac{1}{2} ln (2 x - 1)]}_{1}^{9}$

substituting limits into their integral and subtracting (in any order) (M1)

eg $\frac{π}{2} (ln (17) - ln (1)), π (0 - \frac{1}{2} ln (2 \times 9 - 1))$

correct working involving calculating a log value or using log law (A1)

eg $ln (1) = 0, ln (\frac{17}{1})$

$\frac{π}{2} ln 17 (accept π ln \sqrt{17})$ A1 N3

Note: Full FT may be awarded as normal, from their incorrect answer in part (a), however, do not award the final two A marks unless they involve logarithms.

[4 marks]

Examiners report

[N/A]

18M.1.SL.TZ1.S_7

Question

Consider f(x), g(x) and h(x), for x∈ $R$ where h(x) = $(f \circ g)$ (x).

Given that g(3) = 7 , g′ (3) = 4 and f ′ (7) = −5 , find the gradient of the normal to the curve of h at x = 3.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

recognizing the need to find h′ (M1)

recognizing the need to find h′ (3) (seen anywhere) (M1)

evidence of choosing chain rule (M1)

eg $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}, f^{'} (g (3)) \times g^{'} (3), f^{'} (g) \times g^{'}$

correct working (A1)

eg $f^{'} (7) \times 4, - 5 \times 4$

$h^{'} (3) = - 20$ (A1)

evidence of taking their negative reciprocal for normal (M1)

eg $- \frac{1}{h^{'} (3)}, m_{1} m_{2} = - 1$

gradient of normal is $\frac{1}{20}$ A1 N4

[7 marks]

Examiners report

[N/A]

18M.1.SL.TZ2.S_9a

Question

A closed cylindrical can with radius r centimetres and height h centimetres has a volume of 20 $π$ cm³.

The material for the base and top of the can costs 10 cents per cm² and the material for the curved side costs 8 cents per cm². The total cost of the material, in cents, is C.

Express h in terms of r.

[2]

Show that $C = 20 π r^{2} + \frac{320 π}{r}$ .

[4]

Given that there is a minimum value for C, find this minimum value in terms of $π$ .

[9]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

correct equation for volume (A1)
eg $π r^{2} h = 20 π$

$h = \frac{20}{r^{2}}$ A1 N2

[2 marks]

attempt to find formula for cost of parts (M1)
eg 10 × two circles, 8 × curved side

correct expression for cost of two circles in terms of r (seen anywhere) A1
eg $2 π r^{2} \times 10$

correct expression for cost of curved side (seen anywhere) (A1)
eg $2 π r \times h \times 8$

correct expression for cost of curved side in terms of r A1
eg $8 \times 2 π r \times \frac{20}{r^{2}}, \frac{320 π}{r^{2}}$

$C = 20 π r^{2} + \frac{320 π}{r}$ AG N0

[4 marks]

recognize $C^{'} = 0$ at minimum (R1)
eg $C^{'} = 0, \frac{d C}{d r} = 0$

correct differentiation (may be seen in equation)

$C^{'} = 40 π r - \frac{320 π}{r^{2}}$ A1A1

correct equation A1
eg $40 π r - \frac{320 π}{r^{2}} = 0, 40 π r \frac{320 π}{r^{2}}$

correct working (A1)
eg $40 r^{3} = 320, r^{3} = 8$

r = 2 (m) A1

attempt to substitute their value of r into C
eg $20 π \times 4 + 320 \times \frac{π}{2}$ (M1)

correct working
eg $80 π + 160 π$ (A1)

$240 π$ (cents) A1 N3

Note: Do not accept 753.6, 753.98 or 754, even if 240 $π$ is seen.

[9 marks]

Examiners report

[N/A]

18M.1.SL.TZ1.T_5a

Question

The point A has coordinates (4 , −8) and the point B has coordinates (−2 , 4).

The point D has coordinates (−3 , 1).

Write down the coordinates of C, the midpoint of line segment AB.

[2]

Find the gradient of the line DC.

[2]

Find the equation of the line DC. Write your answer in the form ax + by + d = 0 where a , b and d are integers.

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

(1, −2) (A1)(A1) (C2)
Note: Award (A1) for 1 and (A1) for −2, seen as a coordinate pair.

Accept x = 1, y = −2. Award (A1)(A0) if x and y coordinates are reversed.

[2 marks]

$\frac{1 - (- 2)}{- 3 - 1}$ (M1)

Note: Award (M1) for correct substitution, of their part (a), into gradient formula.

$= - \frac{3}{4} (- 0.75)$ (A1)(ft) (C2)

Note: Follow through from part (a).

[2 marks]

$y - 1 = - \frac{3}{4} (x + 3)$ OR $y + 2 = - \frac{3}{4} (x - 1)$ OR $y = - \frac{3}{4} x - \frac{5}{4}$ (M1)

Note: Award (M1) for correct substitution of their part (b) and a given point.

$1 = - \frac{3}{4} \times - 3 + c$ OR $- 2 = - \frac{3}{4} \times 1 + c$ (M1)

Note: Award (M1) for correct substitution of their part (b) and a given point.

$3 x + 4 y + 5 = 0$ (accept any integer multiple, including negative multiples) (A1)(ft) (C2)

Note: Follow through from parts (a) and (b). Where the gradient in part (b) is found to be $\frac{5}{0}$ , award at most (M1)(A0) for either $x = - 3$ or $x + 3 = 0$ .

[2 marks]

Examiners report

[N/A]

18M.1.SL.TZ2.T_7a

Question

In an international competition, participants can answer questions in only one of the three following languages: Portuguese, Mandarin or Hindi. 80 participants took part in the competition. The number of participants answering in Portuguese, Mandarin or Hindi is shown in the table.

A boy is chosen at random.

State the number of boys who answered questions in Portuguese.

[1]

Find the probability that the boy answered questions in Hindi.

[2]

Two girls are selected at random.

Calculate the probability that one girl answered questions in Mandarin and the other answered questions in Hindi.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

20 (A1) (C1)

[1 mark]

$\frac{5}{43} (0.11627 \dots, 11.6279 \dots %)$ (A1)(A1) (C2)

Note: Award (A1) for correct numerator, (A1) for correct denominator.

[2 marks]

$\frac{7}{37} \times \frac{12}{36} + \frac{12}{37} \times \frac{7}{36}$ (A1)(M1)

Note: Award (A1) for first or second correct product seen, (M1) for adding their two products or for multiplying their product by two.

$= \frac{14}{111} (0.12612 \dots, 12.6126 %)$ (A1) (C3)

[3 marks]

Examiners report

[N/A]

18M.1.SL.TZ2.T_13a

Question

A factory produces shirts. The cost, C, in Fijian dollars (FJD), of producing x shirts can be modelled by

C(x) = (x − 75)² + 100.

The cost of production should not exceed 500 FJD. To do this the factory needs to produce at least 55 shirts and at most s shirts.

Find the cost of producing 70 shirts.

[2]

Find the value of s.

[2]

Find the number of shirts produced when the cost of production is lowest.

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

(70 − 75)² + 100 (M1)

Note: Award (M1) for substituting in x = 70.

125 (A1) (C2)

[2 marks]

(s − 75)² + 100 = 500 (M1)

Note: Award (M1) for equating C(x) to 500. Accept an inequality instead of =.

(M1)

Note: Award (M1) for sketching correct graph(s).

(s =) 95 (A1) (C2)

[2 marks]

(M1)

Note: Award (M1) for an attempt at finding the minimum point using graph.

$\frac{95 + 55}{2}$ (M1)

Note: Award (M1) for attempting to find the mid-point between their part (b) and 55.

(C'(x) =) 2x − 150 = 0 (M1)

Note: Award (M1) for an attempt at differentiation that is correctly equated to zero.

75 (A1) (C2)

[2 marks]

Examiners report

[N/A]

18N.1.SL.TZ0.S_6

Question

Let $f (x) = \frac{6 - 2 x}{\sqrt{16 + 6 x - x^{2}}}$ . The following diagram shows part of the graph of $f$ .

The region R is enclosed by the graph of $f$ , the $x$ -axis, and the $y$ -axis. Find the area of R.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1 (limits in terms of $x$ )

valid approach to find $x$ -intercept (M1)

eg $f (x) = 0$ , $\frac{6 - 2 x}{\sqrt{16 + 6 x - x^{2}}} = 0$ , $6 - 2 x = 0$

$x$ -intercept is 3 (A1)

valid approach using substitution or inspection (M1)

eg $u = 16 + 6 x - x^{2}$ , $\int_{0}^{3} \frac{6 - 2 x}{\sqrt{u}} d x$ , $d u = 6 - 2 x$ , $\int \frac{1}{\sqrt{u}}$ , $2 u^{\frac{1}{2}}$ ,

$u = \sqrt{16 + 6 x - x^{2}}$ , $\frac{d u}{d x} = (6 - 2 x) \frac{1}{2} {(16 + 6 x - x^{2})}^{- \frac{1}{2}}$ , $\int 2 d u$ , $2 u$

$\int f (x) d x = 2 \sqrt{16 + 6 x - x^{2}}$ (A2)

substituting both of their limits into their integrated function and subtracting (M1)

eg $2 \sqrt{16 + 6 (3) - 3^{2}} - 2 \sqrt{16 + 6 {(0)}^{2} - 0^{2}}$ , $2 \sqrt{16 + 18 - 9} - 2 \sqrt{16}$

Note: Award M0 if they substitute into original or differentiated function. Do not accept only “– 0” as evidence of substituting lower limit.

correct working (A1)

eg $2 \sqrt{25} - 2 \sqrt{16}$ , $10 - 8$

area = 2 A1 N2

METHOD 2 (limits in terms of $u$ )

valid approach to find $x$ -intercept (M1)

eg $f (x) = 0$ , $\frac{6 - 2 x}{\sqrt{16 + 6 x - x^{2}}} = 0$ , $6 - 2 x = 0$

$x$ -intercept is 3 (A1)

valid approach using substitution or inspection (M1)

eg $u = 16 + 6 x - x^{2}$ , $\int_{0}^{3} \frac{6 - 2 x}{\sqrt{u}} d x$ , $d u = 6 - 2 x$ , $\int \frac{1}{\sqrt{u}}$ ,

$u = \sqrt{16 + 6 x - x^{2}}$ , $\frac{d u}{d x} = (6 - 2 x) \frac{1}{2} {(16 + 6 x - x^{2})}^{- \frac{1}{2}}$ , $\int 2 d u$

correct integration (A2)

eg $\int \frac{1}{\sqrt{u}} d u = 2 u^{\frac{1}{2}}$ , $\int 2 d u = 2 u$

both correct limits for $u$ (A1)

eg $u$ = 16 and $u$ = 25, $\int_{16}^{25} \frac{1}{\sqrt{u}} d u$ , ${[2 u^{\frac{1}{2}}]}_{16}^{25}$ , $u$ = 4 and $u$ = 5, $\int_{4}^{5} 2 d u$ , ${[2 u]}_{4}^{5}$

substituting both of their limits for $u$ (do not accept 0 and 3) into their integrated function and subtracting (M1)

eg $2 \sqrt{25} - 2 \sqrt{16}$ , $10 - 8$

Note: Award M0 if they substitute into original or differentiated function, or if they have not attempted to find limits for $u$ .

area = 2 A1 N2

[8 marks]

Examiners report

[N/A]

18N.1.SL.TZ0.S_10c

Question

Let $f (x) = x^{3} - 2 x^{2} + a x + 6$ . Part of the graph of $f$ is shown in the following diagram.

The graph of $f$ crosses the $y$ -axis at the point P. The line L is tangent to the graph of $f$ at P.

Find $f^{'} (x)$ .

[2]

b.i.

Hence, find the equation of L in terms of $a$ .

[4]

b.ii.

The graph of $f$ has a local minimum at the point Q. The line L passes through Q.

Find the value of $a$ .

[8]

Markscheme

$f^{'} = 3 x^{2} - 4 x + a$ A2 N2

[2 marks]

b.i.

valid approach (M1)

eg $f^{'} (0)$

correct working (A1)

eg $3 {(0)}^{2} - 4 (0) + a$ , slope = $a$ , $f^{'} (0) = a$

attempt to substitute gradient and coordinates into linear equation (M1)

eg $y - 6 = a (x - 0)$ , $y - 0 = a (x - 6)$ , $6 = a (0) + c$ , L $= a x + 6$

correct equation A1 N3

eg $y = a x + 6$ , $y - 6 = a x$ , $y - 6 = a (x - 0)$

[4 marks]

b.ii.

valid approach to find intersection (M1)

eg $f (x) = L$

correct equation (A1)

eg $x^{3} - 2 x^{2} + a x + 6 = a x + 6$

correct working (A1)

eg $x^{3} - 2 x^{2} = 0$ , $x^{2} (x - 2) = 0$

$x = 2$ at Q (A1)

valid approach to find minimum (M1)

eg $f^{'} (x) = 0$

correct equation (A1)

eg $3 x^{2} - 4 x + a = 0$

substitution of their value of $x$ at Q into their $f^{'} (x) = 0$ equation (M1)

eg $3 {(2)}^{2} - 4 (2) + a = 0$ , $12 - 8 + a = 0$

$a$ = −4 A1 N0

[8 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

18N.1.SL.TZ0.T_11c

Question

Consider the curve y = 5x³− 3x.

The curve has a tangent at the point P(−1, −2).

Find the equation of this tangent. Give your answer in the form y = mx + c.

Markscheme

(y − (−2)) = 12 (x − (−1)) (M1)

−2 = 12(−1) + c (M1)

Note: Award (M1) for point and their gradient substituted into the equation of a line.

y = 12x + 10 (A1)(ft) (C2)

Note: Follow through from part (b).

[2 marks]

Examiners report

[N/A]

19M.1.SL.TZ1.S_5

Question

The derivative of a function $f$ is given by $f^{'} (x) = 2 e^{- 3 x}$ . The graph of $f$ passes through $(\frac{1}{3}, 5)$ .

Find $f (x)$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

recognizing to integrate (M1)

eg $\int f^{'}$ , $\int 2 e^{- 3 x} d x$ , $d u = - 3$

correct integral (do not penalize for missing + $C$ ) (A2)

eg $- \frac{2}{3} e^{- 3 x} + C$

substituting $(\frac{1}{3}, 5)$ (in any order) into their integrated expression (must have + $C$ ) M1

eg $- \frac{2}{3} e^{- 3 (1 / 3)} + C = 5$

Note: Award M0 if they substitute into original or differentiated function.

$f (x) = - \frac{2}{3} e^{- 3 x} + 5 + \frac{2}{3} e^{- 1}$ (or any equivalent form, eg $- \frac{2}{3} e^{- 3 x} + 5 - \frac{2}{- 3 e}$ ) A1 N4

[5 marks]

Examiners report

[N/A]

19M.1.SL.TZ1.S_7a

Question

A particle P starts from point O and moves along a straight line. The graph of its velocity, $v$ ms⁻¹ after $t$ seconds, for 0 ≤ $t$ ≤ 6 , is shown in the following diagram.

The graph of $v$ has $t$ -intercepts when $t$ = 0, 2 and 4.

The function $s (t)$ represents the displacement of P from O after $t$ seconds.

It is known that P travels a distance of 15 metres in the first 2 seconds. It is also known that $s (2) = s (5)$ and $\int_{2}^{4} v d t = 9$ .

Find the value of $s (4) - s (2)$ .

[2]

Find the total distance travelled in the first 5 seconds.

[5]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

recognizing relationship between $v$ and $s$ (M1)

eg $\int v = s$ , $s^{'} = v$

$s (4) - s (2) = 9$ A1 N2

[2 marks]

correctly interpreting distance travelled in first 2 seconds (seen anywhere, including part (a) or the area of 15 indicated on diagram) (A1)

eg $\int_{0}^{2} v = 15$ , $s (2) = 15$

valid approach to find total distance travelled (M1)

eg sum of 3 areas, $\int_{0}^{4} v + \int_{4}^{5} v$ , shaded areas in diagram between 0 and 5

Note: Award M0 if only $\int_{0}^{5} | v |$ is seen.

correct working towards finding distance travelled between 2 and 5 (seen anywhere including within total area expression or on diagram) (A1)

eg $\int_{2}^{4} v - \int_{4}^{5} v$ , $\int_{2}^{4} v = \int_{4}^{5} | v |$ , $\int_{4}^{5} v d t = - 9$ , $s (4) - s (2) - [s (5) - s (4)]$ ,

equal areas

correct working using $s (5) = s (2)$ (A1)

eg $15 + 9 - (- 9)$ , $15 + 2 [s (4) - s (2)]$ , $15 + 2 (9)$ , $2 \times s (4) - s (2)$ , $48 - 15$

total distance travelled = 33 (m) A1 N2

[5 marks]

Examiners report

[N/A]

19M.1.SL.TZ2.S_9a

Question

Let $θ$ be an obtuse angle such that $sin θ = \frac{3}{5}$ .

Let $f (x) = e^{x} sin x - \frac{3 x}{4}$ .

Find the value of $tan θ$ .

[4]

Line $L$ passes through the origin and has a gradient of $tan θ$ . Find the equation of $L$ .

[2]

The following diagram shows the graph of $f$ for 0 ≤ $x$ ≤ 3. Line $M$ is a tangent to the graph of $f$ at point P.

Given that $M$ is parallel to $L$ , find the $x$ -coordinate of P.

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

evidence of valid approach (M1)

eg sketch of triangle with sides 3 and 5, $co s^{2} θ = 1 - si n^{2} θ$

correct working (A1)

eg missing side is 4 (may be seen in sketch), $cos θ = \frac{4}{5}$ , $cos θ = - \frac{4}{5}$

$tan θ = - \frac{3}{4}$ A2 N4

[4 marks]

correct substitution of either gradient or origin into equation of line (A1)

(do not accept $y = m x + b$ )

eg $y = x tan θ$ , $y - 0 = m (x - 0)$ , $y = m x$

$y = - \frac{3}{4} x$ A2 N4

Note: Award A1A0 for $L = - \frac{3}{4} x$ .

[2 marks]

valid approach to equate their gradients (M1)

eg $f^{'} = tan θ$ , $f^{'} = - \frac{3}{4}$ , $e^{x} cos x + e^{x} sin x - \frac{3}{4} = - \frac{3}{4}$ , $e^{x} (cos x + sin x) - \frac{3}{4} = - \frac{3}{4}$

correct equation without $e^{x}$ (A1)

eg $sin x = - cos x$ , $cos x + sin x = 0$ , $\frac{- sin x}{cos x} = 1$

correct working (A1)

eg $tan θ = - 1$ , $x = 135^{\circ}$

$x = \frac{3 π}{4}$ (do not accept $135^{\circ}$ ) A1 N1

Note: Do not award the final A1 if additional answers are given.

[4 marks]

Examiners report

[N/A]

19M.1.SL.TZ2.S_10b

Question

Let $y = {(x^{3} + x)}^{\frac{3}{2}}$ .

Consider the functions $f (x) = \sqrt{x^{3} + x}$ and $g (x) = 6 - 3 x^{2} \sqrt{x^{3} + x}$ , for $x$ ≥ 0.

The graphs of $f$ and $g$ are shown in the following diagram.

The shaded region $R$ is enclosed by the graphs of $f$ , $g$ , the $y$ -axis and $x = 1$ .

Hence find $\int (3 x^{2} + 1) \sqrt{x^{3} + x} d x$ .

[3]

Write down an expression for the area of $R$ .

[2]

Hence find the exact area of $R$ .

[6]

Markscheme

integrating by inspection from (a) or by substitution (M1)

eg $\frac{2}{3} \int \frac{3}{2} (3 x^{2} + 1) \sqrt{x^{3} + x} d x$ , $u = x^{3} + x$ , $\frac{d u}{d x} = 3 x^{2} + 1$ , $\int u^{\frac{1}{2}}$ , $\frac{u^{\frac{3}{2}}}{1.5}$

correct integrated expression in terms of $x$ A2 N3

eg $\frac{2}{3} {(x^{3} + x)}^{\frac{3}{2}} + C$ , $\frac{{(x^{3} + x)}^{1.5}}{1.5} + C$

[3 marks]

integrating and subtracting functions (in any order) (M1)

eg $\int g - f$ , $\int f - \int g$

correct integral (including limits, accept absence of $d x$ ) A1 N2

eg $\int_{0}^{1} (g - f) d x$ , $\int_{0}^{1} 6 - 3 x^{2} \sqrt{x^{3} + x} - \sqrt{x^{3} + x} d x$ , $\int_{0}^{1} g (x) - \int_{0}^{1} f (x)$

[2 marks]

recognizing $\sqrt{x^{3} + x}$ is a common factor (seen anywhere, may be seen in part (c)) (M1)

eg $(- 3 x^{2} - 1) \sqrt{x^{3} + x}$ , $\int 6 - (3 x^{2} + 1) \sqrt{x^{3} + x}$ , $(3 x^{2} - 1) \sqrt{x^{3} + x}$

correct integration (A1)(A1)

eg $6 x - \frac{2}{3} {(x^{3} + x)}^{\frac{3}{2}}$

Note: Award A1 for $6 x$ and award A1 for $- \frac{2}{3} {(x^{3} + x)}^{\frac{3}{2}}$ .

substituting limits into their integrated function and subtracting (in any order) (M1)

eg $6 - \frac{2}{3} {(1^{3} + 1)}^{\frac{3}{2}}$ , $0 - [6 - \frac{2}{3} {(1^{3} + 1)}^{\frac{3}{2}}]$

correct working (A1)

eg $6 - \frac{2}{3} \times 2 \sqrt{2}$ , $6 - \frac{2}{3} \times \sqrt{4} \times \sqrt{2}$

area of $R = 6 - \frac{4 \sqrt{2}}{3} (= 6 - \frac{2}{3} \sqrt{8}, 6 - \frac{2}{3} \times 2^{\frac{3}{2}}, \frac{18 - 4 \sqrt{2}}{3})$ A1 N3

[6 marks]

Examiners report

[N/A]

19M.1.SL.TZ2.T_5a

Question

A school café sells three flavours of smoothies: mango ( $M$ ), kiwi fruit ( $K$ ) and banana ( $B$ ).
85 students were surveyed about which of these three flavours they like.

35 students liked mango, 37 liked banana, and 26 liked kiwi fruit
2 liked all three flavours
20 liked both mango and banana
14 liked mango and kiwi fruit
3 liked banana and kiwi fruit

Using the given information, complete the following Venn diagram.

[2]

Find the number of surveyed students who did not like any of the three flavours.

[2]

A student is chosen at random from the surveyed students.

Find the probability that this student likes kiwi fruit smoothies given that they like mango smoothies.

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

(A1)(A1) (C2)

Note: Award (A1) for 18, 12 and 1 in correct place on Venn diagram, (A1) for 3, 16 and 11 in correct place on Venn diagram.

[2 marks]

85 − (3 + 16 + 11 + 18 + 12 + 1 + 2) (M1)

Note: Award (M1) for subtracting the sum of their values from 85.

22 (A1)(ft) (C2)

Note: Follow through from their Venn diagram in part (a).
If any numbers that are being subtracted are negative award (M1)(A0).

[2 marks]

$\frac{14}{35} (\frac{2}{5}, 0.4, 40 %)$ (A1)(ft)(A1)(ft) (C2)

Note: Award (A1) for correct numerator; (A1) for correct denominator. Follow through from their Venn diagram.

[2 marks]

Examiners report

[N/A]

19M.1.SL.TZ1.T_12a

Question

The diagram shows a circular horizontal board divided into six equal sectors. The sectors are labelled white (W), yellow (Y) and blue (B).

A pointer is pinned to the centre of the board. The pointer is to be spun and when it stops the colour of the sector on which the pointer stops is recorded. The pointer is equally likely to stop on any of the six sectors.

Eva will spin the pointer twice. The following tree diagram shows all the possible outcomes.

Find the probability that both spins are yellow.

[2]

Find the probability that at least one of the spins is yellow.

[3]

Write down the probability that the second spin is yellow, given that the first spin is blue.

[1]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\frac{1}{3} \times \frac{1}{3}$ OR ${(\frac{1}{3})}^{2}$ (M1)

Note: Award (M1) for multiplying correct probabilities.

$\frac{1}{9}$ (0.111, 0.111111…, 11.1%) (A1) (C2)

[2 marks]

$(\frac{1}{2} \times \frac{1}{3}) + (\frac{1}{6} \times \frac{1}{3}) + \frac{1}{3}$ (M1)(M1)

Note: Award (M1) for $(\frac{1}{2} \times \frac{1}{3})$ and $(\frac{1}{6} \times \frac{1}{3})$ or equivalent, and (M1) for $\frac{1}{3}$ and adding only the three correct probabilities.

$1 - {(\frac{2}{3})}^{2}$ (M1)(M1)

Note: Award (M1) for $\frac{2}{3}$ seen and (M1) for subtracting ${(\frac{2}{3})}^{2}$ from 1. This may be shown in a tree diagram with “yellow” and “not yellow” branches.

$\frac{5}{9}$ (0.556, 0.555555…, 55.6%) (A1)(ft) (C3)

Note: Follow through marks may be awarded if their answer to part (a) is used in a correct calculation.

[3 marks]

$\frac{1}{3}$ (0.333, 0.333333…, 33.3%) (A1) (C1)

[1 mark]

Examiners report

[N/A]

19M.1.SL.TZ1.T_15a

Question

A cylinder with radius $r$ and height $h$ is shown in the following diagram.

The sum of $r$ and $h$ for this cylinder is 12 cm.

Write down an equation for the area, $A$ , of the curved surface in terms of $r$ .

[2]

Find $\frac{d A}{d r}$ .

[2]

Find the value of $r$ when the area of the curved surface is maximized.

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$A = 2 π r (12 - r)$ OR $A = 24 π r - 2 π r^{2}$ (A1)(M1) (C2)

Note: Award (A1) for $r + h = 12$ or $h = 12 - r$ seen. Award (M1) for correctly substituting into curved surface area of a cylinder. Accept $A = 2 π r (12 - r)$ OR $A = 24 π r - 2 π r^{2}$ .

[2 marks]

$24 π - 4 π r$ (A1)(ft)(A1)(ft) (C2)

Note: Award (A1)(ft) for $24 π$ and (A1)(ft) for $- 4 π r$ . Follow through from part (a). Award at most (A1)(ft)(A0) if additional terms are seen.

[2 marks]

$24 π - 4 π r = 0$ (M1)

Note: Award (M1) for setting their part (b) equal to zero.

6 (cm) (A1)(ft) (C2)

Note: Follow through from part (b).

[2 marks]

Examiners report

[N/A]

19M.1.SL.TZ2.T_15a

Question

A potter sells $x$ vases per month.

His monthly profit in Australian dollars (AUD) can be modelled by

$P (x) = - \frac{1}{5} x^{3} + 7 x^{2} - 120, x ⩾ 0.$

Find the value of $P$ if no vases are sold.

[1]

Differentiate $P (x)$ .

[2]

Hence, find the number of vases that will maximize the profit.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

−120 (AUD) (A1) (C1)

[1 mark]

$- \frac{3}{5} x^{2} + 14 x$ (A1)(A1) (C2)

Note: Award (A1) for each correct term. Award at most (A1)(A0) for extra terms seen.

[2 marks]

$- \frac{3}{5} x^{2} + 14 x = 0$ (M1)

Note: Award (M1) for equating their derivative to zero.

sketch of their derivative (approximately correct shape) with $x$ -intercept seen (M1)

$23 \frac{1}{3} (23.3, 23.3333 \dots, \frac{70}{3})$ (A1)(ft)

Note: Award (C2) for $23 \frac{1}{3} (23.3, 23.3333 \dots, \frac{70}{3})$ seen without working.

23 (A1)(ft) (C3)

Note: Follow through from part (b).

[3 marks]

Examiners report

[N/A]

19N.1.SL.TZ0.S_8a

Question

A small cuboid box has a rectangular base of length $3 x$ cm and width $x$ cm, where $x > 0$ . The height is $y$ cm, where $y > 0$ .

The sum of the length, width and height is $12$ cm.

The volume of the box is $V$ cm³.

Write down an expression for $y$ in terms of $x$ .

[1]

Find an expression for $V$ in terms of $x$ .

[2]

Find $\frac{d V}{d x}$ .

[2]

Find the value of $x$ for which $V$ is a maximum.

[4]

d.i.

Justify your answer.

[3]

d.ii.

Find the maximum volume.

[2]

Markscheme

$y = 12 - 4 x$ A1 N1

[1 mark]

correct substitution into volume formula (A1)

eg $3 x \times x \times y, x \times 3 x \times (12 - x - 3 x), (12 - 4 x) (x) (3 x)$

$V = 3 x^{2} (12 - 4 x) (= 36 x^{2} - 12 x^{3})$ A1 N2

Note: Award A0 for unfinished answers such as $3 x^{2} (12 - x - 3 x)$ .

[2 marks]

$\frac{d V}{d x} = 72 x - 36 x^{2}$ A1A1 N2

Note: Award A1 for $72 x$ and A1 for $- 36 x^{2}$ .

[2 marks]

valid approach to find maximum (M1)

eg $V^{'} = 0, 72 x - 36 x^{2} = 0$

correct working (A1)

eg $x (72 - 36 x), \frac{- 72 \pm \sqrt{72^{2} - 4 \cdot (- 36) \cdot 0}}{2 (- 36)}, 36 x = 72, 36 x (2 - x) = 0$

$x = 2$ A2 N2

Note: Award A1 for $x = 2$ and $x = 0$ .

[4 marks]

d.i.

valid approach to explain that $V$ is maximum when $x = 2$ (M1)

eg attempt to find $V^{″}$ , sign chart (must be labelled $V^{'}$ )

correct value/s A1

eg $V^{″} (2) = 72 - 72 \times 2$ , $V^{'} (a)$ where $a < 2$ and $V^{'} (b)$ where $b > 2$

correct reasoning R1

eg $V^{″} (2) < 0$ , $V^{'}$ is positive for $x < 2$ and negative for $x > 2$

Note: Do not award R1 unless A1 has been awarded.

$V$ is maximum when $x = 2$ AG N0

[3 marks]

d.ii.

correct substitution into their expression for volume A1

eg $3 \times 2^{2} (12 - 4 \times 2)$ , $36 (2^{2}) - 12 (2^{3})$

$V = 48$ (cm³) A1 N1

[2 marks]

Examiners report

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

19N.1.SL.TZ0.S_10a

Question

Let $g (x) = p^{x} + q$ , for $x, p, q \in R, p > 1$ . The point $A (0, a)$ lies on the graph of $g$ .

Let $f (x) = g^{- 1} (x)$ . The point $B$ lies on the graph of $f$ and is the reflection of point $A$ in the line $y = x$ .

The line $L_{1}$ is tangent to the graph of $f$ at $B$ .

Write down the coordinates of $B$ .

[2]

Given that $f^{'} (a) = \frac{1}{ln p}$ , find the equation of $L_{1}$ in terms of $x$ , $p$ and $q$ .

[5]

The line $L_{2}$ is tangent to the graph of $g$ at $A$ and has equation $y = (ln p) x + q + 1$ .

The line $L_{2}$ passes through the point $(- 2, - 2)$ .

The gradient of the normal to $g$ at $A$ is $\frac{1}{ln (\frac{1}{3})}$ .

Find the equation of $L_{1}$ in terms of $x$ .

[7]

Markscheme

$B (a, 0)$ (accept $B (q + 1, 0)$ ) A2 N2

[2 marks]

Note: There are many approaches to this part, and the steps may be done in any order. Please check working and award marks in line with the markscheme, noting that candidates may work with the equation of the line before finding $a$ .

FINDING $a$

valid attempt to find an expression for $a$ in terms of $q$ (M1)

$g (0) = a, p^{0} + q = a$

$a = q + 1$ (A1)

FINDING THE EQUATION OF $L_{1}$

EITHER

attempt to substitute tangent gradient and coordinates into equation of straight line (M1)

eg $y - 0 = f^{'} (a) (x - a), y = f^{'} (a) (x - (q + 1))$

correct equation in terms of $a$ and $p$ (A1)

eg $y - 0 = \frac{1}{ln (p)} (x - a)$

attempt to substitute tangent gradient and coordinates to find $b$

eg $0 = \frac{1}{ln (p)} (a) + b$

$b = \frac{- a}{ln (p)}$ (A1)

THEN (must be in terms of both $p$ and $q$ )

$y = \frac{1}{ln p} (x - q - 1), y = \frac{1}{ln p} x - \frac{q + 1}{ln p}$ A1 N3

Note: Award A0 for final answers in the form $L_{1} = \frac{1}{ln p} (x - q - 1)$

[5 marks]

Note: There are many approaches to this part, and the steps may be done in any order. Please check working and award marks in line with the markscheme, noting that candidates may find $q$ in terms of $p$ before finding a value for $p$ .

FINDING $p$

valid approach to find the gradient of the tangent (M1)

eg $m_{1} m_{2} = - 1, - \frac{1}{\frac{1}{ln (\frac{1}{3})}}, - ln (\frac{1}{3}), - \frac{1}{ln p} = \frac{1}{ln (\frac{1}{3})}$

correct application of log rule (seen anywhere) (A1)

eg $ln {(\frac{1}{3})}^{- 1}, - (ln (1) - ln (3))$

correct equation (seen anywhere) A1

eg $ln p = ln 3, p = 3$

FINDING $q$

correct substitution of $(- 2, - 2)$ into $L_{2}$ equation (A1)

eg $- 2 = (ln p) (- 2) + q + 1$

$q = 2 ln p - 3, q = 2 ln 3 - 3$ (seen anywhere) A1

FINDING $L_{1}$

correct substitution of their $p$ and $q$ into their $L_{1}$ (A1)

eg $y = \frac{1}{ln 3} (x - (2 ln 3 - 3) - 1)$

$y = \frac{1}{ln 3} (x - 2 ln 3 + 2), y = \frac{1}{ln 3} x - \frac{2 ln 3 - 2}{ln 3}$ A1 N2

Note: Award A0 for final answers in the form $L_{1} = \frac{1}{ln 3} (x - 2 ln 3 + 2)$ .

[7 marks]

Examiners report

[N/A]

20N.1.SL.TZ0.S_3a

Question

Let $f (x) = \sqrt{12 - 2 x}, x \leq a$ . The following diagram shows part of the graph of $f$ .

The shaded region is enclosed by the graph of $f$ , the $x$ -axis and the $y$ -axis.

The graph of $f$ intersects the $x$ -axis at the point $(a, 0)$ .

Find the value of $a$ .

[2]

Find the volume of the solid formed when the shaded region is revolved $360 °$ about the $x$ -axis.

[5]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

recognize $f (x) = 0$ (M1)

eg $\sqrt{12 - 2 x} = 0, 2 x = 12$

$a = 6$ (accept $x = 6$ , $(6, 0)$ ) A1 N2

[2 marks]

attempt to substitute either their limits or the function into volume formula (must involve $f^{2}$ ) (M1)

eg $\int_{0}^{6} f^{2} d x, π \int {(\sqrt{12 - 2 x})}^{2}, π \int_{0}^{6} 12 - 2 x d x$

correct integration of each term A1 A1

eg $12 x - x^{2}, 12 x - x^{2} + c, {[12 x - x^{2}]}_{0}^{6}$

substituting limits into their integrated function and subtracting (in any order) (M1)

eg $π (12 (6) - {(6)}^{2}) - π (0), 72 π - 36 π, (12 (6) - {(6)}^{2}) - (0)$

Note: Award M0 if candidate has substituted into $f, f^{2}$ or $f'$ .

volume $= 36 π$ A1 N2

[5 marks]

Examiners report

[N/A]

20N.1.SL.TZ0.S_6

Question

The graph of a function $f$ passes through the point $(\ln 4, 20)$ .

Given that $f' (x) = 6 e^{2 x}$ , find $f (x)$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

evidence of integration (M1)

eg $\int f' (x) d x, \int 6 e^{2 x}$

correct integration (accept missing $+ c$ ) (A1)

eg $\frac{1}{2} \times 6 e^{2 x}, 3 e^{2 x} + c$

substituting initial condition into their integrated expression (must have $+ c$ ) M1

eg $3 e^{2 \times \ln 4} + c = 20$

Note: Award M0 if candidate has substituted into $f'$ or $f''$ .

correct application of $\log (a^{b}) = b \log a$ rule (seen anywhere) (A1)

eg $2 \ln 4 = \ln 16, e^{\ln 16}, \ln 4^{2}$

correct application of $e^{\ln a} = a$ rule (seen anywhere) (A1)

eg $e^{\ln 16} = 16, {(e^{\ln 4})}^{2} = 4^{2}$

correct working (A1)

eg $3 \times 16 + c = 20, 3 \times {(4)}^{2} + c = 20, c = - 28$

$f (x) = 3 e^{2 x} - 28$ A1 N4

[7 marks]

Examiners report

[N/A]

20N.1.SL.TZ0.S_7a

Question

In this question, all lengths are in metres and time is in seconds.

Consider two particles, $P_{1}$ and $P_{2}$ , which start to move at the same time.

Particle $P_{1}$ moves in a straight line such that its displacement from a fixed-point is given by $s (t) = 10 - \frac{7}{4} t^{2}$ , for $t \geq 0$ .

Find an expression for the velocity of $P_{1}$ at time $t$ .

[2]

Particle $P_{2}$ also moves in a straight line. The position of $P_{2}$ is given by $r = (\begin{matrix} - 1 \\ 6 \end{matrix}) + t (\begin{matrix} 4 \\ - 3 \end{matrix})$ .

The speed of $P_{1}$ is greater than the speed of $P_{2}$ when $t > q$ .

Find the value of $q$ .

[5]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

recognizing velocity is derivative of displacement (M1)

eg $v = \frac{d s}{d t}, \frac{d}{d t} (10 - \frac{7}{4} t^{2})$

velocity $= - \frac{14}{4} t (= - \frac{7}{2} t)$ A1 N2

[2 marks]

valid approach to find speed of $P_{2}$ (M1)

eg $|(\begin{matrix} 4 \\ - 3 \end{matrix})|, \sqrt{4^{2} + {(- 3)}^{2}}$ , velocity $= \sqrt{4^{2} + {(- 3)}^{2}}$

correct speed (A1)

eg $5 {m s}^{- 1}$

recognizing relationship between speed and velocity (may be seen in inequality/equation) R1

eg $|- \frac{7}{2} t|$ , speed = | velocity | , graph of $P_{1}$ speed , $P_{1}$ speed $= \frac{7}{2} t, P_{2}$ velocity $= - 5$

correct inequality or equation that compares speed or velocity (accept any variable for $q$ ) A1

eg $|- \frac{7}{2} t| > 5, - \frac{7}{2} q < - 5, \frac{7}{2} q > 5, \frac{7}{2} q = 5$

$q = \frac{10}{7}$ (seconds) (accept $t > \frac{10}{7}$ , do not accept $t = \frac{10}{7}$ ) A1 N2

Note: Do not award the last two A1 marks without the R1.

[5 marks]

Examiners report

[N/A]

20N.1.SL.TZ0.S_10b

Question

The following diagram shows part of the graph of $f (x) = \frac{k}{x}$ , for $x > 0, k > 0$ .

Let $P (p, \frac{k}{p})$ be any point on the graph of $f$ . Line $L_{1}$ is the tangent to the graph of $f$ at $P$ .

Line $L_{1}$ intersects the $x$ -axis at point $A (2 p, 0)$ and the $y$ -axis at point B.

Find $f' (p)$ in terms of $k$ and $p$ .

[2]

a.i.

Show that the equation of $L_{1}$ is $k x + p^{2} y - 2 p k = 0$ .

[2]

a.ii.

Find the area of triangle $AOB$ in terms of $k$ .

[5]

The graph of $f$ is translated by $(\begin{matrix} 4 \\ 3 \end{matrix})$ to give the graph of $g$ .
In the following diagram:

point $Q$ lies on the graph of $g$
points $C$ , $D$ and $E$ lie on the vertical asymptote of $g$
points $D$ and $F$ lie on the horizontal asymptote of $g$
point $G$ lies on the $x$ -axis such that $FG$ is parallel to $DC$ .

Line $L_{2}$ is the tangent to the graph of $g$ at $Q$ , and passes through $E$ and $F$ .

Given that triangle $EDF$ and rectangle $CDFG$ have equal areas, find the gradient of $L_{2}$ in terms of $p$ .

[6]

Markscheme

$f' (x) = - k x^{- 2}$ (A1)

$f' (p) = - k p^{- 2} (= - \frac{k}{p^{2}})$ A1 N2

[2 marks]

a.i.

attempt to use point and gradient to find equation of $L_{1}$ M1

eg $y - \frac{k}{p} = - k p^{- 2} (x - p), \frac{k}{p} = - \frac{k}{p^{2}} (p) + b$

correct working leading to answer A1

eg $p^{2} y - k p = - k x + k p, y - \frac{k}{p} = - \frac{k}{p^{2}} x + \frac{k}{p}, y = - \frac{k}{p^{2}} x + \frac{2 k}{p}$

$k x + p^{2} y - 2 p k = 0$ AG N0

[2 marks]

a.ii.

METHOD 1 – area of a triangle

recognizing $x = 0$ at $B$ (M1)

correct working to find $y$ -coordinate of null (A1)

eg $p^{2} y - 2 p k = 0$

$y$ -coordinate of null at $y = \frac{2 k}{p}$ (may be seen in area formula) A1

correct substitution to find area of triangle (A1)

eg $\frac{1}{2} (2 p) (\frac{2 k}{p}), p \times (\frac{2 k}{p})$

area of triangle $AOB = 2 k$ A1 N3

METHOD 2 – integration

recognizing to integrate $L_{1}$ between $0$ and $2 p$ (M1)

eg $\int_{0}^{2 p} L_{1} d x, \int_{0}^{2 p} - \frac{k}{p^{2}} x + \frac{2 k}{p}$

correct integration of both terms A1

eg $- \frac{k x^{2}}{2 p^{2}} + \frac{2 k x}{p}, - \frac{k}{2 p^{2}} x^{2} + \frac{2 k}{p} x + c, {[- \frac{k}{2 p^{2}} x^{2} + \frac{2 k}{p} x]}_{0}^{2 p}$

substituting limits into their integrated function and subtracting (in either order) (M1)

eg $- \frac{k {(2 p)}^{2}}{2 p^{2}} + \frac{2 k (2 p)}{p} - (0), - \frac{4 k p^{2}}{2 p^{2}} + \frac{4 k p}{p}$

correct working (A1)

eg $- 2 k + 4 k$

area of triangle $AOB = 2 k$ A1 N3

[5 marks]

Note: In this question, the second M mark may be awarded independently of the other marks, so it is possible to award (M0)(A0)M1(A0)(A0)A0.

recognizing use of transformation (M1)

eg area of triangle $AOB$ = area of triangle $DEF$ , $g (x) = \frac{k}{x - 4} + 3,$ gradient of $L_{2} =$ gradient of $L_{1}$ , $D (4, 3),2p+4,$ one correct shift

correct working (A1)

eg area of triangle $DEF = 2 k, CD = 3, DF = 2 p, CG = 2 p, E (4, \frac{2 k}{p} + 3), F (2 p + 4, 3), Q (p + 4, \frac{k}{p} + 3),$

gradient of $L_{2} = - \frac{k}{p^{2}}, g' (x) = - \frac{k}{{(x - 4)}^{2}},$ area of rectangle $CDFG = 2 k$

valid approach (M1)

eg $\frac{ED \times DF}{2} = CD \times DF, 2 p \cdot 3 = 2 k, ED = 2 CD, \int_{4}^{2 p + 4} L_{2} d x = 4 k$

correct working (A1)

eg $ED = 6, E (4, 9), k = 3 p, gradient = \frac{3 - (\frac{2 k}{p} + 3)}{(2 p + 4) - 4}, \frac{- 6}{(\frac{2 k}{3})}, - \frac{9}{k}$

correct expression for gradient (in terms of $p$ ) (A1)

eg $\frac{- 6}{2 p}, \frac{9 - 3}{4 - (2 p + 4)}, - \frac{3 p}{p^{2}}, \frac{3 - (\frac{2 (3 p)}{p} + 3)}{(2 p + 4) - 4}, - \frac{9}{3 p}$

gradient of $L_{2}$ is $- \frac{3}{p} (= - 3 p^{- 1})$ A1 N3

[6 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

20N.1.SL.TZ0.T_13a

Question

Consider the graph of the function $f (x) = x^{2} - \frac{k}{x}$ .

The equation of the tangent to the graph of $y = f (x)$ at $x = - 2$ is $2 y = 4 - 5 x$ .

Write down $f' (x)$ .

[3]

Write down the gradient of this tangent.

[1]

Find the value of $k$ .

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure. It appeared in a paper that permitted the use of a calculator, and so might not be suitable for all forms of practice.

$2 x + \frac{k}{x^{2}}$ (A1)(A1)(A1) (C3)

Note: Award (A1) for $2 x$ , (A1) for $+ k$ , and (A1) for $x^{- 2}$ or $\frac{1}{x^{2}}$ .
Award at most (A1)(A1)(A0) if additional terms are seen.

[3 marks]

$- 2.5 (\frac{- 5}{2})$ (A1) (C1)

[1 mark]

$- 2.5 = 2 \times (- 2) + \frac{k}{{(- 2)}^{2}}$ (M1)

Note: Award (M1) for equating their gradient from part (b) to their substituted derivative from part (a).

$(k =) 6$ (A1)(ft) (C2)

Note: Follow through from parts (a) and (b).

[2 marks]

Examiners report

[N/A]

21M.1.SL.TZ1.5a

Question

Consider the functions $f (x) = - {(x - h)}^{2} + 2 k$ and $g (x) = e^{x - 2} + k$ where $h, k \in ℝ$ .

The graphs of $f$ and $g$ have a common tangent at $x = 3$ .

Find $f' (x)$ .

[1]

Show that $h = \frac{e + 6}{2}$ .

[3]

Hence, show that $k = e + \frac{e^{2}}{4}$ .

[3]

Markscheme

$f' (x) = - 2 (x - h)$ A1

[1 mark]

$g' (x) = e^{x - 2}$ OR $g' (3) = e^{3 - 2}$ (may be seen anywhere) A1

Note: The derivative of $g$ must be explicitly seen, either in terms of $x$ or $3$ .

recognizing $f' (3) = g' (3)$ (M1)

$- 2 (3 - h) = e^{3 - 2} (= e)$

$- 6 + 2 h = e$ OR $3 - h = - \frac{e}{2}$ A1

Note: The final A1 is dependent on one of the previous marks being awarded.

$h = \frac{e + 6}{2}$ AG

[3 marks]

$f (3) = g (3)$ (M1)

$- {(3 - h)}^{2} + 2 k = e^{3 - 2} + k$

correct equation in $k$

EITHER

$- {(3 - \frac{e + 6}{2})}^{2} + 2 k = e^{3 - 2} + k$ A1

$k = e + {(\frac{6 - e - 6}{2})}^{2} (= e + {(\frac{- e}{2})}^{2})$ A1

$k = e + {(3 - \frac{e + 6}{2})}^{2}$ A1

$k = e + 9 - 3 e - 18 + \frac{e^{2} + 12 e + 36}{4}$ A1

THEN

$k = e + \frac{e^{2}}{4}$ AG

[3 marks]

Examiners report

[N/A]

21M.1.SL.TZ2.5b

Question

Consider the function $f$ defined by $f (x) = \ln (x^{2} - 16)$ for $x > 4$ .

The following diagram shows part of the graph of $f$ which crosses the $x$ -axis at point $A$ , with coordinates $(a, 0)$ . The line $L$ is the tangent to the graph of $f$ at the point $B$ .

Find the exact value of $a$ .

[3]

Given that the gradient of $L$ is $\frac{1}{3}$ , find the $x$ -coordinate of $B$ .

[6]

Markscheme

$\ln (x^{2} - 16) = 0$ (M1)

$e^{0} = x^{2} - 16 (= 1)$

$x^{2} = 17$ OR $x = \pm \sqrt{17}$ (A1)

$a = \sqrt{17}$ A1

[3 marks]

attempt to differentiate (must include $2 x$ and/or $\frac{1}{x^{2} - 16}$ ) (M1)

$f' (x) = \frac{2 x}{x^{2} - 16}$ A1

setting their derivative $= \frac{1}{3}$ M1

$\frac{2 x}{x^{2} - 16} = \frac{1}{3}$

$x^{2} - 16 = 6 x$ OR $x^{2} - 6 x - 16 = 0$ (or equivalent) A1

valid attempt to solve their quadratic (M1)

$x = 8$ A1

Note: Award A0 if the candidate’s final answer includes additional solutions (such as $x = - 2, 8$ ).

[6 marks]

Examiners report

[N/A]

21M.1.SL.TZ1.8a

Question

Let $y = \frac{\ln x}{x^{4}}$ for $x > 0$ .

Consider the function defined by $f (x) \frac{\ln x}{x^{4}}$ for $x > 0$ and its graph $y = f (x)$ .

Show that $\frac{d y}{d x} = \frac{1 - 4 \ln x}{x^{5}}$ .

[3]

The graph of $f$ has a horizontal tangent at point $P$ . Find the coordinates of $P$ .

[5]

Given that $f'' (x) = \frac{20 \ln x - 9}{x^{6}}$ , show that $P$ is a local maximum point.

[3]

Solve $f (x) > 0$ for $x > 0$ .

[2]

Sketch the graph of $f$ , showing clearly the value of the $x$ -intercept and the approximate position of point $P$ .

[3]

Markscheme

attempt to use quotient or product rule (M1)

$\frac{d y}{d x} = \frac{x^{4} (\frac{1}{x}) - (\ln x) (4 x^{3})}{{(x^{4})}^{2}}$ OR $(\ln x) (- 4 x^{- 5}) + (x^{- 4}) (\frac{1}{x})$ A1

correct working A1

$= \frac{x^{3} (1 - 4 \ln x)}{x^{8}}$ OR cancelling $x^{3}$ OR $\frac{- 4 \ln x}{x^{5}} + \frac{1}{x^{5}}$

$= \frac{1 - 4 \ln x}{x^{5}}$ AG

[3 marks]

$f' (x) = \frac{d y}{d x} = 0$ (M1)

$\frac{1 - 4 \ln x}{x^{5}} = 0$

$\ln x = \frac{1}{4}$ (A1)

$x = e^{\frac{1}{4}}$ A1

substitution of their $x$ to find $y$ (M1)

$y = \frac{\ln e^{\frac{1}{4}}}{{(e^{\frac{1}{4}})}^{4}}$

$= \frac{1}{4 e} (= \frac{1}{4} e^{- 1})$ A1

$P (e^{\frac{1}{4}}, \frac{1}{4 e})$

[5 marks]

$f'' (e^{\frac{1}{4}}) = \frac{20 \ln e^{\frac{1}{4}} - 9}{{(e^{\frac{1}{4}})}^{6}}$ (M1)

$= \frac{5 - 9}{e^{1.5}} (= - \frac{4}{e^{1.5}})$ A1

which is negative R1

hence $P$ is a local maximum AG

Note: The R1 is dependent on the previous A1 being awarded.

[3 marks]

$\ln x > 0$ (A1)

$x > 1$ A1

[2 marks]

A1A1A1

Note: Award A1 for one $x$ -intercept only, located at $1$

A1 for local maximum, $P$ , in approximately correct position
A1 for curve approaching $x$ -axis as $x \to \infty$ (including change in concavity).

[3 marks]

Examiners report

[N/A]

21M.1.SL.TZ2.8b

Question

Consider the function $f$ defined by $f (x) = 6 + 6 \cos x$ , for $0 \leq x \leq 4 π$ .

The following diagram shows the graph of $y = f (x)$ .

The graph of $f$ touches the $x$ -axis at points $A$ and $B$ , as shown. The shaded region is enclosed by the graph of $y = f (x)$ and the $x$ -axis, between the points $A$ and $B$ .

The right cone in the following diagram has a total surface area of $12 π$ , equal to the shaded area in the previous diagram.

The cone has a base radius of $2$ , height $h$ , and slant height $l$ .

Find the $x$ -coordinates of $A$ and $B$ .

[3]

Show that the area of the shaded region is $12 π$ .

[5]

Find the value of $l$ .

[3]

Hence, find the volume of the cone.

[4]

Markscheme

$6 + 6 \cos x = 0$ (or setting their $f' (x) = 0$ ) (M1)

$\cos x = - 1$ (or $\sin x = 0$ )

$x = π, x = 3 π$ A1A1

[3 marks]

attempt to integrate $\int_{π}^{3 π} (6 + 6 \cos x) d x$ (M1)

$= {[6 x + 6 \sin x]}_{π}^{3 π}$ A1A1

substitute their limits into their integrated expression and subtract (M1)

$= (18 π + 6 \sin 3 π) - (6 π + 6 \sin π)$

$= (6 (3 π) + 0) - (6 π + 0) (= 18 π - 6 π)$ A1

area $= 12 π$ AG

[5 marks]

attempt to substitute into formula for surface area (including base) (M1)

$π (2^{2}) + π (2) (l) = 12 π$ (A1)

$4 π + 2 π l = 12 π$

$2 π l = 8 π$

$l = 4$ A1

[3 marks]

valid attempt to find the height of the cone (M1)

e.g. $2^{2} + h^{2} = {(their l)}^{2}$

$h = \sqrt{12} (= 2 \sqrt{3})$ (A1)

attempt to use $V = \frac{1}{3} π r^{2} h$ with their values substituted M1

$(\frac{1}{3} π (2^{2}) (\sqrt{12}))$

$volume = \frac{4 π \sqrt{12}}{3} (= \frac{8 π \sqrt{3}}{3} = \frac{8 π}{\sqrt{3}})$ A1

[4 marks]

Examiners report

[N/A]

21M.1.SL.TZ2.9a

Question

Particle A travels in a straight line such that its displacement, $s$ metres, from a fixed origin after $t$ seconds is given by $s (t) = 8 t - t^{2}$ , for $0 \leq t \leq 10$ , as shown in the following diagram.

Particle A starts at the origin and passes through the origin again when $t = p$ .

Particle A changes direction when $t = q$ .

The total distance travelled by particle A is given by $d$ .

Find the value of $p$ .

[2]

Find the value of $q$ .

[2]

b.i.

Find the displacement of particle A from the origin when $t = q$ .

[2]

b.ii.

Find the distance of particle A from the origin when $t = 10$ .

[2]

Find the value of $d$ .

[2]

A second particle, particle B, travels along the same straight line such that its velocity is given by $v (t) = 14 - 2 t$ , for $t \geq 0$ .

When $t = k$ , the distance travelled by particle B is equal to $d$ .

Find the value of $k$ .

[4]

Markscheme

setting $s (t) = 0$ (M1)

$8 t - t^{2} = 0$

$t (8 - t) = 0$

$p = 8$ (accept $t = 8, (8, 0)$ ) A1

Note: Award A0 if the candidate’s final answer includes additional solutions (such as $p = 0, 8$ ).

[2 marks]

recognition that when particle changes direction $v = 0$ OR local maximum on graph of $s$ OR vertex of parabola (M1)

$q = 4$ (accept $t = 4$ ) A1

[2 marks]

b.i.

substituting their value of $q$ into $s (t)$ OR integrating $v (t)$ from $t = 0$ to $t = 4$ (M1)

$displacement = 16 (m)$ A1

[2 marks]

b.ii.

$s (10) = - 20$ OR $distance = |s (t)|$ OR integrating $v (t)$ from $t = 0$ to $t = 10$ (M1)

$distance = 20 (m)$ A1

[2 marks]

$16$ forward $+ 36$ backward OR $16 + 16 + 20$ OR $\int_{0}^{10} |v (t)| d t$ (M1)

$d = 52 (m)$ A1

[2 marks]

METHOD 1

graphical method with triangles on $v (t)$ graph M1

$49 + (\frac{x (2 x)}{2})$ (A1)

$49 + x^{2} = 52, x = \sqrt{3}$ (A1)

$k = 7 + \sqrt{3}$ A1

METHOD 2

recognition that distance $= \int |v (t)| d t$ M1

$\int_{0}^{7} (14 - 2 t) d t + \int_{7}^{k} (2 t - 14) d t$

${[14 t - t^{2}]}_{0}^{7} + {[t^{2} - 14 t]}_{7}^{k}$ (A1)

$14 (7) - 7^{2} + ((k^{2} - 14 k) - (7^{2} - 14 (7))) = 52$ (A1)

$k = 7 + \sqrt{3}$ A1

[4 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

21N.1.SL.TZ0.2

Question

Given that $\frac{d y}{d x} = \cos (x - \frac{π}{4})$ and $y = 2$ when $x = \frac{3 π}{4}$ , find $y$ in terms of $x$ .

Markscheme

METHOD 1

recognition that $y = \int \cos (x - \frac{π}{4}) d x$ (M1)

$y = \sin (x - \frac{π}{4}) (+ c)$ (A1)

substitute both $x$ and $y$ values into their integrated expression including $c$ (M1)

$2 = \sin \frac{π}{2} + c$

$c = 1$

$y = \sin (x - \frac{π}{4}) + 1$ A1

METHOD 2

$\int_{2}^{y} d y = \int_{\frac{3 π}{4}}^{x} \cos (x - \frac{π}{4}) d x$ (M1)(A1)

$y - 2 = \sin (x - \frac{π}{4}) - \sin \frac{π}{2}$ A1

$y = \sin (x - \frac{π}{4}) + 1$ A1

[4 marks]

Examiners report

[N/A]

21N.1.SL.TZ0.5a

Question

The function $f$ is defined for all $x \in ℝ$ . The line with equation $y = 6 x - 1$ is the tangent to the graph of $f$ at $x = 4$ .

The function $g$ is defined for all $x \in ℝ$ where $g (x) = x^{2} - 3 x$ and $h (x) = f (g (x))$ .

Write down the value of $f' (4)$ .

[1]

Find $f (4)$ .

[1]

Find $h (4)$ .

[2]

Hence find the equation of the tangent to the graph of $h$ at $x = 4$ .

[3]

Markscheme

$f' (4) = 6$ A1

[1 mark]

$f (4) = 6 \times 4 - 1 = 23$ A1

[1 mark]

$h (4) = f (g (4))$ (M1)

$h (4) = f (4^{2} - 3 \times 4) = f (4)$

$h (4) = 23$ A1

[2 marks]

attempt to use chain rule to find $h'$ (M1)

$f' (g (x)) \times g' (x)$ OR $(x^{2} - 3 x)' \times f' (x^{2} - 3 x)$

$h' (4) = (2 \times 4 - 3) f' (4^{2} - 3 \times 4)$ A1

$= 30$

$y - 23 = 30 (x - 4)$ OR $y = 30 x - 97$ A1

[3 marks]

Examiners report

[N/A]

21N.1.SL.TZ0.7b

Question

A particle $P$ moves along the $x$ -axis. The velocity of $P$ is $v m s^{- 1}$ at time $t$ seconds, where $v (t) = 4 + 4 t - 3 t^{2}$ for $0 \leq t \leq 3$ . When $t = 0, P$ is at the origin $O$ .

Find the value of $t$ when $P$ reaches its maximum velocity.

[2]

a.i.

Show that the distance of $P$ from $O$ at this time is $\frac{88}{27}$ metres.

[5]

a.ii.

Sketch a graph of $v$ against $t$ , clearly showing any points of intersection with the axes.

[4]

Find the total distance travelled by $P$ .

[5]

Markscheme

valid approach to find turning point ( $v' = 0, - \frac{b}{2 a}$ , average of roots) (M1)

$4 - 6 t = 0$ OR $- \frac{4}{2 (- 3)}$ OR $\frac{- \frac{2}{3} + 2}{2}$

$t = \frac{2}{3}$ (s) A1

[2 marks]

a.i.

attempt to integrate $v$ (M1)

$\int v d t = \int (4 + 4 t - 3 t^{2}) d t = 4 t + 2 t^{2} - t^{3} (+ c)$ A1A1

Note: Award A1 for $4 t + 2 t^{2}$ , A1 for $- t^{3}$ .

attempt to substitute their $t$ into their solution for the integral (M1)

distance $= 4 (\frac{2}{3}) + 2 {(\frac{2}{3})}^{2} - {(\frac{2}{3})}^{3}$

$= \frac{8}{3} + \frac{8}{9} - \frac{8}{27}$ (or equivalent) A1

$= \frac{88}{27}$ (m) AG

[5 marks]

a.ii.

valid approach to solve $4 + 4 t - 3 t^{2} = 0$ (may be seen in part (a)) (M1)

$(2 - t) (2 + 3 t)$ OR $\frac{- 4 \pm \sqrt{16 + 48}}{- 6}$

correct $x$ - intercept on the graph at $t = 2$ A1

Note: The following two A marks may only be awarded if the shape is a concave down parabola. These two marks are independent of each other and the (M1).

correct domain from $0$ to $3$ starting at $(0, 4)$ A1

Note: The $3$ must be clearly indicated.

vertex in approximately correct place for $t = \frac{2}{3}$ and $v > 4$ A1

[4 marks]

recognising to integrate between $0$ and $2$ , or $2$ and $3$ OR $\int_{0}^{3} |4 + 4 t - 3 t^{2}| d t$ (M1)

$\int_{0}^{2} (4 + 4 t - 3 t^{2}) d t$

$= 8$ A1

$\int_{2}^{3} (4 + 4 t - 3 t^{2}) d t$

$= - 5$ A1

valid approach to sum the two areas (seen anywhere) (M1)

$\int_{0}^{2} v d t - \int_{2}^{3} v d t$ OR $\int_{0}^{2} v d t + |\int_{2}^{3} v d t|$

total distance travelled $= 13$ (m) A1

[5 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

21N.1.SL.TZ0.9a

Question

Consider a function $f$ with domain $a < x < b$ . The following diagram shows the graph of $f'$ , the derivative of $f$ .

The graph of $f'$ , the derivative of $f$ , has $x$ -intercepts at $x = p, x = 0$ and $x = t$ . There are local maximum points at $x = q$ and $x = t$ and a local minimum point at $x = r$ .

Find all the values of $x$ where the graph of $f$ is increasing. Justify your answer.

[2]

Find the value of $x$ where the graph of $f$ has a local maximum.

[1]

Find the value of $x$ where the graph of $f$ has a local minimum. Justify your answer.

[2]

c.i.

Find the values of $x$ where the graph of $f$ has points of inflexion. Justify your answer.

[3]

c.ii.

The total area of the region enclosed by the graph of $f'$ , the derivative of $f$ , and the $x$ -axis is $20$ .

Given that $f (p) + f (t) = 4$ , find the value of $f (0)$ .

[6]

Markscheme

Special note: In this question if candidates use the word 'gradient' in their reasoning. e.g. gradient is positive, it must be clear whether this is the gradient of $f$ or the gradient of $f'$ to earn the R mark.

$f$ increases when $p < x < 0$ A1

$f$ increases when $f' (x) > 0$ OR $f'$ is above the $x$ -axis R1

Note: Do not award A0R1.

[2 marks]

$x = 0$ A1

[1 mark]

$f$ is minimum when $x = p$ A1

because $f' (p) = 0, f' (x) < 0$ when $x < p$ and $f' (x) > 0$ when $x > p$

(may be seen in a sign diagram clearly labelled as $f'$ )

OR because $f'$ changes from negative to positive at $x = p$

OR $f' (p) = 0$ and slope of $f'$ is positive at $x = p$ R1

Note: Do not award A0 R1

[2 marks]

c.i.

$f$ has points of inflexion when $x = q, x = r$ and $x = t$ A2

$f'$ has turning points at $x = q, x = r$ and $x = t$

$f'' (q) = 0, f'' (r) = 0$ and $f'' (t) = 0$ and $f'$ changes from increasing to decreasing or vice versa at each of these $x$ -values (may be seen in a sign diagram clearly labelled as $f''$ and $f'$ ) R1

Note: Award A0 if any incorrect answers are given. Do not award A0R1

[3 marks]

c.ii.

recognizing area from $p$ to $t$ (seen anywhere) M1

$\int_{p}^{t} |f' (x)| d x$

recognizing to negate integral for area below $x$ -axis (M1)

$\int_{p}^{0} f' (x) d x - \int_{0}^{t} f' (x) d x$ OR $\int_{p}^{0} f' (x) d x + \int_{t}^{0} f' (x) d x$

$\int_{m}^{n} f' (x) d x = f (n) - f (m)$ (for any integral) (M1)

$f (0) - f (p) - [f (t) - f (0)]$ OR $f (0) - f (p) + f (0) - f (t)$ (A1)

$2 f (0) - [f (t) + f (p)] = 20, 2 f (0) - 4 = 20$ (A1)

$f (0) = 12$ A1

[6 marks]

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

22M.1.SL.TZ1.2b

Question

The expression $\frac{3 \sqrt{x} - 5}{\sqrt{x}}$ can be written as $3 - 5 x^{p}$ . Write down the value of $p$ .

[1]

Hence, find the value of $\int_{1}^{9} (\frac{3 \sqrt{x} - 5}{\sqrt{x}}) d x$ .

[4]

Markscheme

$\frac{3 \sqrt{x} - 5}{\sqrt{x}} = 3 - 5 x^{- \frac{1}{2}}$ A1

$p = - \frac{1}{2}$

[1 mark]

$\int \frac{3 \sqrt{x} - 5}{\sqrt{x}} d x = 3 x - 10 x^{\frac{1}{2}} (+ c)$ A1A1

substituting limits into their integrated function and subtracting (M1)

$3 (9) - 10 {(9)}^{\frac{1}{2}} - (3 (1) - 10 {(1)}^{\frac{1}{2}})$ OR $27 - 10 \times 3 - (3 - 10)$

$= 4$ A1

[4 marks]

Examiners report

Many candidates could give the value of p correctly. However, many did struggle with the integration, including substituting limits into the integrand, without integrating at all. An incorrect value of p often resulted in arithmetic of greater complexity.

[N/A]

22M.1.SL.TZ1.5

Question

Consider the curve with equation $y = (2 x - 1) e^{k x}$ , where $x \in ℝ$ and $k \in ℚ$ .

The tangent to the curve at the point where $x = 1$ is parallel to the line $y = 5 e^{k} x$ .

Find the value of $k$ .

Markscheme

evidence of using product rule (M1)

$\frac{d y}{d x} = (2 x - 1) \times (k e^{k x}) + 2 \times e^{k x} (= e^{k x} (2 k x - k + 2))$ A1

correct working for one of (seen anywhere) A1

$\frac{d y}{d x}$ at $x = 1 \Rightarrow k e^{k} + 2 e^{k}$

slope of tangent is $5 e^{k}$

their $\frac{d y}{d x}$ at $x = 1$ equals the slope of $y = 5 e^{k} x (= 5 e^{k})$ (seen anywhere) (M1)

$k e^{k} + 2 e^{k} = 5 e^{k}$

$k = 3$ A1

[5 marks]

Examiners report

The product rule was well recognised and used with 𝑥=1 properly substituted into this expression. Although the majority of the candidates tried equating the derivative to the slope of the tangent line, the slope of the tangent line was not correctly identified; many candidates incorrectly substituted 𝑥=1 into the tangent equation, thus finding the y-coordinate instead of the slope.

22M.1.SL.TZ2.7c

Question

The following diagram shows part of the graph of a quadratic function $f$ .

The graph of $f$ has its vertex at $(3, 4)$ , and it passes through point $Q$ as shown.

The function can be written in the form $f (x) = a {(x - h)}^{2} + k$ .

The line $L$ is tangent to the graph of $f$ at $Q$ .

Now consider another function $y = g (x)$ . The derivative of $g$ is given by $g' (x) = f (x) - d$ , where $d \in ℝ$ .

Write down the equation of the axis of symmetry.

[1]

Write down the values of $h$ and $k$ .

[2]

b.i.

Point $Q$ has coordinates $(5, 12)$ . Find the value of $a$ .

[2]

b.ii.

Find the equation of $L$ .

[4]

Find the values of $d$ for which $g$ is an increasing function.

[3]

Find the values of $x$ for which the graph of $g$ is concave-up.

[3]

Markscheme

$x = 3$ A1

Note: Must be an equation in the form “ $x =$ ”. Do not accept $3$ or $\frac{- b}{2 a} = 3$ .

[1 mark]

$h = 3, k = 4$ (accept $a {(x - 3)}^{2} + 4$ ) A1A1

[2 marks]

b.i.

attempt to substitute coordinates of $Q$ (M1)

$12 = a {(5 - 3)}^{2} + 4, 4 a + 4 = 12$

$a = 2$ A1

[2 marks]

b.ii.

recognize need to find derivative of $f$ (M1)

$f' (x) = 4 (x - 3)$ or $f' (x) = 4 x - 12$ A1

$f' (5) = 8$ (may be seen as gradient in their equation) (A1)

$y - 12 = 8 (x - 5)$ or $y = 8 x - 28$ A1

Note: Award A0 for $L = 8 x - 28$ .

[4 marks]

METHOD 1

Recognizing that for $g$ to be increasing, $f (x) - d > 0$ , or $g' > 0$ (M1)

The vertex must be above the $x$ -axis, $4 - d > 0, d - 4 < 0$ (R1)

$d < 4$ A1

METHOD 2

attempting to find discriminant of $g'$ (M1)

${(- 12)}^{2} - 4 (2) (22 - d)$

recognizing discriminant must be negative (R1)

$- 32 + 8 d < 0$ OR $Δ < 0$

$d < 4$ A1

[3 marks]

recognizing that for $g$ to be concave up, $g'' > 0$ (M1)

$g'' > 0$ when $f' > 0, 4 x - 12 > 0, x - 3 > 0$ (R1)

$x > 3$ A1

[3 marks]

Examiners report

In parts (a) and (b) of this question, a majority of candidates recognized the connection between the coordinates of the vertex and the axis of symmetry and the values of $h$ and $k$ , and most candidates were able to successfully substitute the coordinates of point Q to find the value of $a$ . In part (c), the candidates who recognized the need to use the derivative to find the gradient of the tangent were generally successful in finding the equation of the line, although many did not give their equation in the proper form in terms of $x$ and $y$ , and instead wrote $L = 8 x - 28$ , thus losing the final mark. Parts (d) and (e) were much more challenging for candidates. Although a good number of candidates recognized that $g' (x) > 0$ in part (d), and $g'' (x) > 0$ in part (e), very few were able to proceed beyond this point and find the correct inequalities for their final answers.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

22M.1.SL.TZ2.8b

Question

Consider the functions $f (x) = \frac{1}{x - 4} + 1$ , for $x \neq 4$ , and $g (x) = x - 3$ for $x \in ℝ$ .

The following diagram shows the graphs of $f$ and $g$ .

The graphs of $f$ and $g$ intersect at points $A$ and $B$ . The coordinates of $A$ are $(3, 0)$ .

In the following diagram, the shaded region is enclosed by the graph of $f$ , the graph of $g$ , the $x$ -axis, and the line $x = k$ , where $k \in ℤ$ .

The area of the shaded region can be written as $\ln (p) + 8$ , where $p \in ℤ$ .

Find the coordinates of $B$ .

[5]

Find the value of $k$ and the value of $p$ .

[10]

Markscheme

$\frac{1}{x - 4} + 1 = x - 3$ (M1)

$x^{2} - 8 x + 15 = 0$ OR ${(x - 4)}^{2} = 1$ (A1)

valid attempt to solve their quadratic (M1)

$(x - 3) (x - 5) = 0$ OR $x = \frac{8 \pm \sqrt{8^{2} - 4 (1) (15)}}{2 (1)}$ OR $(x - 4) = \pm 1$

$x = 5 (x = 3, x = 5)$ (may be seen in answer) A1

$B (5, 2)$ (accept $x = 5, y = 2$ ) A1

[5 marks]

recognizing two correct regions from $x = 3$ to $x = 5$ and from $x = 5$ to $x = k$ (R1)

triangle $+ \int_{5}^{k} f (x) d x$ OR $\int_{3}^{5} g (x) d x + \int_{5}^{k} f (x) d x$ OR $\int_{3}^{5} (x - 3) d x + \int_{5}^{k} (\frac{1}{x - 4} + 1) d x$

area of triangle is $2$ OR $\frac{2 \cdot 2}{2}$ OR $(\frac{5^{2}}{2} - 3 (5)) - (\frac{3^{2}}{2} - 3 (3))$ (A1)

correct integration (A1)(A1)

$\int (\frac{1}{x - 4} + 1) d x = \ln (x - 4) + x (+ C)$

Note: Award A1 for $\ln (x - 4)$ and A1 for $x$ .
Note: The first three A marks may be awarded independently of the R mark.

substitution of their limits (for $x$ ) into their integrated function (in terms of $x$ ) (M1)

$\ln (k - 4) + k - (\ln 1 + 5)$

${[\ln (x - 4) + x]}_{5}^{k} = \ln (k - 4) + k - 5$ A1

adding their two areas (in terms of $k$ ) and equating to $\ln p + 8$ (M1)

$2 + \ln (k - 4) + k - 5 = \ln p + 8$

equating their non-log terms to $8$ (equation must be in terms of $k$ ) (M1)

$k - 3 = 8$

$k = 11$ A1

$11 - 4 = p$

$p = 7$ A1

[10 marks]

Examiners report

Nearly all candidates knew to set up an equation with $f (x) = g (x)$ in order to find the intersection of the two graphs, and most were able to solve the resulting quadratic equation. Candidates were not as successful in part (b), however. While some candidates recognized that there were two regions to be added together, very few were able to determine the correct boundaries of these regions, with many candidates integrating one or both functions from $x = 3$ to $x = k$ . While a good number of candidates were able to correctly integrate the function(s), without the correct bounds the values of $k$ and $p$ were unattainable.

[N/A]

22M.1.SL.TZ1.7e.i

Question

A function, $f$ , has its derivative given by $f' (x) = 3 x^{2} - 12 x + p$ , where $p \in ℝ$ . The following diagram shows part of the graph of $f'$ .

The graph of $f'$ has an axis of symmetry $x = q$ .

The vertex of the graph of $f'$ lies on the $x$ -axis.

The graph of $f$ has a point of inflexion at $x = a$ .

Find the value of $q$ .

[2]

Write down the value of the discriminant of $f'$ .

[1]

b.i.

Hence or otherwise, find the value of $p$ .

[3]

b.ii.

Find the value of the gradient of the graph of $f'$ at $x = 0$ .

[3]

Sketch the graph of $f ″$ , the second derivative of $f$ . Indicate clearly the $x$ -intercept and the $y$ -intercept.

[2]

Write down the value of $a$ .

[1]

e.i.

Find the values of $x$ for which the graph of $f$ is concave-down. Justify your answer.

[2]

e.ii.

Markscheme

EITHER

attempt to use $x = - \frac{b}{2 a}$ (M1)

$q = - \frac{- 12}{2 \times 3}$

attempt to complete the square (M1)

$3 {(x - 2)}^{2} - 12 + p$

attempt to differentiate and equate to $0$ (M1)

$f'' (x) = 6 x - 12 = 0$

THEN

$q = 2$ A1

[2 marks]

discriminant $= 0$ A1

[1 mark]

b.i.

EITHER

attempt to substitute into $b^{2} - 4 a c$ (M1)

${(- 12)}^{2} - 4 \times 3 \times p = 0$ A1

$f' (2) = 0$ (M1)

$- 12 + p = 0$ A1

THEN

$p = 12$ A1

[3 marks]

b.ii.

$f'' (x) = 6 x - 12$ A1

attempt to find $f'' (0)$ (M1)

$= 6 \times 0 - 12$

gradient $= - 12$ A1

[3 marks]

A1A1

Note: Award A1 for line with positive gradient, A1 for correct intercepts.

[2 marks]

$a = 2$ A1

[1 mark]

e.i.

$x < 2$ A1

$f'' (x) < 0$ (for $x < 2$ ) OR the $f''$ is below the $x$ -axis (for $x < 2$ )

OR $f''$ (sign diagram must be labelled $f''$ ) R1

[2 marks]

e.ii.

Examiners report

Candidates did score well on this question. As always, candidates are encouraged to read the questions carefully for key words such as 'value' as opposed to 'expression'. So, if asked for the value of the discriminant, their answer should be a number and not an expression found from $b^{2} - 4 a c$ . As such the value of the discriminant in (b)(i) was often seen in (b)(ii). Please ask students to use a straight edge when sketching a straight line! Overall, the reasoning mark for determining where the graph of f is concave-down, was an improvement on previous years. Sign diagrams were typically well labelled, and the description contained clarity regarding which function was being referred to.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

e.i.

[N/A]

e.ii.

22M.1.SL.TZ1.9b.i

Question

Consider $f (x) = 4 \cos x (1 - 3 \cos 2 x + 3 \cos^{2} 2 x - \cos^{3} 2 x)$ .

Expand and simplify ${(1 - a)}^{3}$ in ascending powers of $a$ .

[2]

a.i.

By using a suitable substitution for $a$ , show that $1 - 3 \cos 2 x + 3 \cos^{2} 2 x - \cos^{3} 2 x = 8 \sin^{6} x$ .

[4]

a.ii.

Show that $\int_{0}^{m} f (x) d x = \frac{32}{7} \sin^{7} m$ , where $m$ is a positive real constant.

[4]

b.i.

It is given that $\int_{m}^{\frac{π}{2}} f (x) d x = \frac{127}{28}$ , where $0 \leq m \leq \frac{π}{2}$ . Find the value of $m$ .

[5]

b.ii.

Markscheme

EITHER

attempt to use binomial expansion (M1)

$1 + {}_{3}C_{1} \times 1 \times (- a) + {}_{3}C_{2} \times 1 \times {(- a)}^{2} + 1 \times {(- a)}^{3}$

$(1 - a) (1 - a) (1 - a)$

$= (1 - a) (1 - 2 a + a^{2})$ (M1)

THEN

$= 1 - 3 a + 3 a^{2} - a^{3}$ A1

[2 marks]

a.i.

$a = \cos 2 x$ (A1)

So, $1 - 3 \cos 2 x + 3 \cos^{2} 2 x - \cos^{3} 2 x =$

${(1 - \cos 2 x)}^{3}$ A1

attempt to substitute any double angle rule for $\cos 2 x$ into ${(1 - \cos 2 x)}^{3}$ (M1)

$= {(2 \sin^{2} x)}^{3}$ A1

$= 8 \sin^{6} x$ AG

Note: Allow working RHS to LHS.

[4 marks]

a.ii.

recognizing to integrate $\int (4 \cos x \times 8 \sin^{6} x) d x$ (M1)

EITHER

applies integration by inspection (M1)

$32 \int (\cos x \times {(\sin x)}^{6}) d x$

$= \frac{32}{7} \sin^{7} x (+ c)$ A1

${[\frac{32}{7} \sin^{7} x]}_{0}^{m} (= \frac{32}{7} \sin^{7} m - \frac{32}{7} \sin^{7} 0)$ A1

$u = \sin x \Rightarrow \frac{d u}{d x} = \cos x$ (M1)

$\int 32 \cos x (\sin^{6} x) d x = \int 32 u^{6} d u$

$= \frac{32}{7} u^{7} (+ c)$ A1

${[\frac{32}{7} \sin^{7} x]}_{0}^{m}$ OR ${[\frac{32}{7} u^{7}]}_{0}^{\sin m} (= \frac{32}{7} \sin^{7} m - \frac{32}{7} \sin^{7} 0)$ A1

THEN

$= \frac{32}{7} \sin^{7} m$ AG

[4 marks]

b.i.

EITHER

$\int_{m}^{\frac{π}{2}} f (x) d x (= {[\frac{32}{7} \sin^{7} x]}_{m}^{\frac{π}{2}}) = \frac{32}{7} \sin^{7} \frac{π}{2} - \frac{32}{7} \sin^{7} m$ M1

$\frac{32}{7} \sin^{7} \frac{π}{2} - \frac{32}{7} \sin^{7} m = \frac{127}{28}$ OR $\frac{32}{7} (1 - \sin^{7} m) = \frac{127}{28}$ (M1)

$\int_{0}^{\frac{π}{2}} f (x) d x = \int_{0}^{m} f (x) d x + \int_{m}^{\frac{π}{2}} f (x) d x$ M1

$\frac{32}{7} = \frac{32}{7} \sin^{7} m + \frac{127}{28}$ (M1)

THEN

$\sin^{7} m = \frac{1}{128} (= \frac{1}{2^{7}})$ (A1)

$\sin m = \frac{1}{2}$ (A1)

$m = \frac{π}{6}$ A1

[5 marks]

b.ii.

Examiners report

Many candidates successfully expanded the binomial, with the most common error being to omit the negative sign with a. The connection between (a)(i) and (ii) was often noted but not fully utilised with candidates embarking on unnecessary complex algebraic expansions of expressions involving double angle rules. Candidates often struggled to apply inspection or substitution when integrating. As a 'show that' question, b(i) provided a useful result to be utilised in (ii). So even without successfully completing (i) candidates could apply it in part (ii). Not many managed to do so.

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

EXN.1.SL.TZ0.1

Question

The derivative of a function $f$ is given by $f' (x) = 3 \sqrt{x}$ .

Given that $f (1) = 3$ , find the value of $f (4)$ .

Markscheme

correct working (A1)

eg −5 + (8 − 1)(3)

u₈ = 16 A1 N2

[2 marks]

METHOD 1

$f (x) = \int 3 \sqrt{x} d x$ (A1)

attempts to integrate (M1)

$f (x) = 2 x^{\frac{3}{2}} + C (= 2 x \sqrt{x} + C)$ A1

uses $f (1) = 3$ to obtain $3 = 2 {(1)}^{\frac{3}{2}} + C$ and so $C = 1$ M1

substitutes $x = 4$ into their expression for $f (x)$ (M1)

so $f (4) = 17$ A1

METHOD 2

$\int_{1}^{4} f' (x) d x = \int_{1}^{4} 3 \sqrt{x} d x$ (A1)

attempts to integrate both sides (M1)

${[f (x)]}_{1}^{4} = {[2 x^{\frac{3}{2}}]}_{1}^{4}$ A1

$f (4) - f (1) = 16 - 2$ M1

uses $f (1) = 3$ to find their value of $f (4)$ (M1)

$f (4) - 3 = 16 - 2$

so $f (4) = 17$ A1

[6 marks]

Examiners report

[N/A]

EXN.1.SL.TZ0.7a

Question

The following diagram shows the graph of $y = 4 - x^{2}$ , $0 \leq x \leq 2$ and rectangle $ORST$ . The rectangle has a vertex at the origin $O$ , a vertex on the $y$ -axis at the point $R (0, y)$ , a vertex on the $x$ -axis at the point $T (x, 0)$ and a vertex at point $S (x, y)$ on the graph.

Let $P$ represent the perimeter of rectangle $ORST$ .

Let $A$ represent the area of rectangle $ORST$ .

Show that $P = - 2 x^{2} + 2 x + 8$ .

[2]

Find the dimensions of rectangle $ORST$ that has maximum perimeter and determine the value of the maximum perimeter.

[6]

Find an expression for $A$ in terms of $x$ .

[2]

Find the dimensions of rectangle $ORST$ that has maximum area.

[5]

Determine the maximum area of rectangle $ORST$ .

[1]

Markscheme

$P = 2 x + 2 y$ (A1)

$= 2 x + 2 (4 - x^{2})$ A1

so $P = - 2 x^{2} + 2 x + 8$ AG

[2 marks]

METHOD 1

EITHER

uses the axis of symmetry of a quadratic (M1)

$x = - \frac{2}{2 (- 2)}$

forms $\frac{d P}{d x} = 0$ (M1)

$- 4 x + 2 = 0$

THEN

$x = \frac{1}{2}$ A1

substitutes their value of $x$ into $y = 4 - x^{2}$ (M1)

$y = 4 - {(\frac{1}{2})}^{2}$

$y = \frac{15}{4}$ A1

so the dimensions of rectangle $ORST$ of maximum perimeter are $\frac{1}{2}$ by $\frac{15}{4}$

EITHER

substitutes their value of $x$ into $P = - 2 x^{2} + 2 x + 8$ (M1)

$P = - 2 {(\frac{1}{2})}^{2} + 2 (\frac{1}{2}) + 8$

substitutes their values of $x$ and $y$ into $P = 2 x + 2 y$ (M1)

$P = 2 (\frac{1}{2}) + 2 (\frac{15}{4})$

$P = \frac{17}{2}$ A1

so the maximum perimeter is $\frac{17}{2}$

METHOD 2

attempts to complete the square M1

$P = - 2 {(x - \frac{1}{2})}^{2} + \frac{17}{2}$ A1

$x = \frac{1}{2}$ A1

substitutes their value of $x$ into $y = 4 - x^{2}$ (M1)

$y = 4 - {(\frac{1}{2})}^{2}$

$y = \frac{15}{4}$ A1

so the dimensions of rectangle $ORST$ of maximum perimeter are $\frac{1}{2}$ by $\frac{15}{4}$

$P = \frac{17}{2}$ A1

so the maximum perimeter is $\frac{17}{2}$

[6 marks]

substitutes $y = 4 - x^{2}$ into $A = x y$ (M1)

$A = x (4 - x^{2}) (= 4 x - x^{3})$ A1

[2 marks]

$\frac{d A}{d x} = 4 - 3 x^{2}$ A1

attempts to solve their $\frac{d A}{d x} = 0$ for $x$ (M1)

$4 - 3 x^{2} = 0$

$\Rightarrow x = \frac{2}{\sqrt{3}} (= \frac{2 \sqrt{3}}{3}) (x > 0)$ A1

substitutes their (positive) value of $x$ into $y = 4 - x^{2}$ (M1)

$y = 4 - {(\frac{2}{\sqrt{3}})}^{2}$

$y = \frac{8}{3}$ A1

[5 marks]

$A = \frac{16}{3 \sqrt{3}} (= \frac{16 \sqrt{3}}{9})$ A1

[1 mark]

Examiners report

[N/A]

EXN.1.SL.TZ0.9e

Question

The following diagram shows a ball attached to the end of a spring, which is suspended from a ceiling.

The height, $h$ metres, of the ball above the ground at time $t$ seconds after being released can be modelled by the function $h (t) = 0.4 \cos (π t) + 1.8$ where $t \geq 0$ .

Find the height of the ball above the ground when it is released.

[2]

Find the minimum height of the ball above the ground.

[2]

Show that the ball takes $2$ seconds to return to its initial height above the ground for the first time.

[2]

For the first 2 seconds of its motion, determine the amount of time that the ball is less than $1.8 + 0.2 \sqrt{2}$ metres above the ground.

[5]

Find the rate of change of the ball’s height above the ground when $t = \frac{1}{3}$ . Give your answer in the form $p π \sqrt{q} {ms}^{- 1}$ where $p \in ℚ$ and $q \in ℤ^{+}$ .

[4]

Markscheme

attempts to find $h (0)$ (M1)

$h (0) = 0.4 \cos (0) + 1.8 (= 2.2)$

$2.2$ (m) (above the ground) A1

[2 marks]

EITHER

uses the minimum value of $\cos (π t)$ which is $- 1$ M1

$0.4 (- 1) + 1.8$ (m)

the amplitude of motion is $0.4$ (m) and the mean position is $1.8$ (m) M1

finds $h' (t) = - 0.4 π \sin (π t)$ , attempts to solve $h' (t) = 0$ for $t$ and determines that the minimum height above the ground occurs at $t = 1, 3, \dots$ M1

$0.4 (- 1) + 1.8$ (m)

THEN

$1.4$ (m) (above the ground) A1

[2 marks]

EITHER

the ball is released from its maximum height and returns there a period later R1

the period is $\frac{2 π}{π} (= 2) (s)$ A1

attempts to solve $h (t) = 2.2$ for $t$ M1

$\cos (π t) = 1$

$t = 0, 2, \dots$ A1

THEN

so it takes $2$ seconds for the ball to return to its initial position for the first time AG

[2 marks]

$0.4 \cos (π t) + 1.8 = 1.8 + 0.2 \sqrt{2}$ (M1)

$0.4 \cos (π t) = 0.2 \sqrt{2}$

$\cos (π t) = \frac{\sqrt{2}}{2}$ A1

$π t = \frac{π}{4}, \frac{7 π}{4}$ (A1)

Note: Accept extra correct positive solutions for $π t$ .

$t = \frac{1}{4}, \frac{7}{4} (0 \leq t \leq 2)$ A1

Note: Do not award A1 if solutions outside $0 \leq t \leq 2$ are also stated.

the ball is less than $1.8 + 0.2 \sqrt{2}$ metres above the ground for $\frac{7}{4} - \frac{1}{4}$ (s)

$1.5$ (s) A1

[5 marks]

EITHER

attempts to find $h' (t)$ (M1)

recognizes that $h' (t)$ is required (M1)

THEN

$h' (t) = - 0.4 π \sin (π t)$ A1

attempts to evaluate their $h' (\frac{1}{3})$ (M1)

$h' (\frac{1}{3}) = - 0.4 π \sin \frac{π}{3}$

$= 0.2 π \sqrt{3} ({ms}^{- 1})$ A1

Note: Accept equivalent correct answer forms where $p \in ℚ$ . For example, $- \frac{1}{5} π \sqrt{3}$ .

[4 marks]

Examiners report

[N/A]

SPM.1.SL.TZ0.4

Question

Let $f^{'} (x) = \frac{8 x}{\sqrt{2 x^{2} + 1}}$ . Given that $f (0) = 5$ , find $f (x)$ .

Markscheme

attempt to integrate (M1)

$u = 2 x^{2} + 1 \Rightarrow \frac{d u}{d x} = 4 x$

$\int \frac{8 x}{\sqrt{2 x^{2} + 1}} d x = \int \frac{2}{\sqrt{u}} d u$ (A1)

EITHER

$= 4 \sqrt{u} (+ C)$ A1

$= 4 \sqrt{2 x^{2} + 1} (+ C)$ A1

THEN

correct substitution into their integrated function (must have C) (M1)

$5 = 4 + C \Rightarrow C = 1$

$f (x) = 4 \sqrt{2 x^{2} + 1} + 1$ A1

[5 marks]

Examiners report

[N/A]

SPM.1.SL.TZ0.8a

Question

Let $f (x) = \frac{1}{3} x^{3} + x^{2} - 15 x + 17$ .

The graph of $f$ has horizontal tangents at the points where $x$ = $a$ and $x$ = $b$ , $a$ < $b$ .

Find $f^{'} (x)$ .

[2]

Find the value of $a$ and the value of $b$ .

[3]

Sketch the graph of $y = f^{'} (x)$ .

[1]

c.i.

Hence explain why the graph of $f$ has a local maximum point at $x = a$ .

[1]

c.ii.

Find $f^{″} (b)$ .

[3]

d.i.

Hence, use your answer to part (d)(i) to show that the graph of $f$ has a local minimum point at $x = b$ .

[1]

d.ii.

The normal to the graph of $f$ at $x = a$ and the tangent to the graph of $f$ at $x = b$ intersect at the point ( $p$ , $q$ ) .

Find the value of $p$ and the value of $q$ .

[5]

Markscheme

$f^{'} (x) = x^{2} + 2 x - 15$ (M1)A1

[2 marks]

correct reasoning that $f^{'} (x) = 0$ (seen anywhere) (M1)

$x^{2} + 2 x - 15 = 0$

valid approach to solve quadratic M1

$(x - 3) (x + 5)$ , quadratic formula

correct values for $x$

3, −5

correct values for $a$ and $b$

$a$ = −5 and $b$ = 3 A1

[3 marks]

[1 mark]

c.i.

first derivative changes from positive to negative at $x = a$ A1

so local maximum at $x = a$ AG

[1 mark]

c.ii.

$f^{″} (x) = 2 x + 2$ A1

substituting their $b$ into their second derivative (M1)

$f^{″} (3) = 2 \times 3 + 2$

$f^{″} (b) = 8$ (A1)

[3 marks]

d.i.

$f^{″} (b)$ is positive so graph is concave up R1

so local minimum at $x = b$ AG

[1 mark]

d.ii.

normal to $f$ at $x = a$ is $x$ = −5 (seen anywhere) (A1)

attempt to find $y$ -coordinate at their value of $b$ (M1)

$f (3) =$ −10 (A1)

tangent at $x = b$ has equation $y$ = −10 (seen anywhere) A1

intersection at (−5, −10)

$p$ = −5 and $q$ = −10 A1

[5 marks]

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

SPM.1.SL.TZ0.9a

Question

Let $f (x) = \frac{ln 5 x}{k x}$ where $x > 0$ , $k \in R^{+}$ .

The graph of $f$ has exactly one maximum point P.

The second derivative of $f$ is given by $f^{″} (x) = \frac{2 ln 5 x - 3}{k x^{3}}$ . The graph of $f$ has exactly one point of inflexion Q.

Show that $f^{'} (x) = \frac{1 - ln 5 x}{k x^{2}}$ .

[3]

Find the x-coordinate of P.

[3]

Show that the x-coordinate of Q is $\frac{1}{5} e^{\frac{3}{2}}$ .

[3]

The region R is enclosed by the graph of $f$ , the x-axis, and the vertical lines through the maximum point P and the point of inflexion Q.

Given that the area of R is 3, find the value of $k$ .

[7]

Markscheme

attempt to use quotient rule (M1)

correct substitution into quotient rule

$f^{'} (x) = \frac{5 k x (\frac{1}{5 x}) - k ln 5 x}{{(k x)}^{2}}$ (or equivalent) A1

$= \frac{k - k ln 5 x}{k^{2} x^{2}}$ , $(k \in R^{+})$ A1

$= \frac{1 - ln 5 x}{k x^{2}}$ AG

[3 marks]

$f^{'} (x) = 0$ M1

$\frac{1 - ln 5 x}{k x^{2}} = 0$

$ln 5 x = 1$ (A1)

$x = \frac{e}{5}$ A1

[3 marks]

$f^{″} (x) = 0$ M1

$\frac{2 ln 5 x - 3}{k x^{3}} = 0$

$ln 5 x = \frac{3}{2}$ A1

$5 x = e^{\frac{3}{2}}$ A1

so the point of inflexion occurs at $x = \frac{1}{5} e^{\frac{3}{2}}$ AG

[3 marks]

attempt to integrate (M1)

$u = ln 5 x \Rightarrow \frac{d u}{d x} = \frac{1}{x}$

$\int \frac{ln 5 x}{k x} d x = \frac{1}{k} \int u d u$ (A1)

EITHER

= $\frac{u^{2}}{2 k}$ A1

so $\frac{1}{k} \int_{1}^{\frac{3}{2}} u d u = {[\frac{u^{2}}{2 k}]}_{1}^{\frac{3}{2}}$ A1

$= \frac{{(ln 5 x)}^{2}}{2 k}$ A1

so $\int_{\frac{e}{5}}^{\frac{1}{5} e^{\frac{3}{2}}} \frac{ln 5 x}{k x} d x = {[\frac{{(ln 5 x)}^{2}}{2 k}]}_{\frac{e}{5}}^{\frac{1}{5} e^{\frac{3}{2}}}$ A1

THEN

$= \frac{1}{2 k} (\frac{9}{4} - 1)$

$= \frac{5}{8 k}$ A1

setting their expression for area equal to 3 M1

$\frac{5}{8 k} = 3$

$k = \frac{5}{24}$ A1

[7 marks]

Examiners report

[N/A]

16N.2.SL.TZ0.S_4a

Question

Let $f (x) = x e^{- x}$ and $g (x) = - 3 f (x) + 1$ .

The graphs of $f$ and $g$ intersect at $x = p$ and $x = q$ , where $p < q$ .

Find the value of $p$ and of $q$ .

[3]

Hence, find the area of the region enclosed by the graphs of $f$ and $g$ .

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid attempt to find the intersection (M1)

eg $f = g$ , sketch, one correct answer

$p = 0.357402, q = 2.15329$

$p = 0.357, q = 2.15$ A1A1 N3

[3 marks]

attempt to set up an integral involving subtraction (in any order) (M1)

eg $\int_{p}^{q} [f (x) - g (x)] d x, \int_{p}^{q} f (x) d x - \int_{p}^{q} g (x) d x$

0.537667

$area = 0.538$ A2 N3

[3 marks]

Examiners report

[N/A]

16N.2.SL.TZ0.S_6a

Question

All lengths in this question are in metres.

Let $f (x) = - 0.8 x^{2} + 0.5$ , for $- 0.5 ⩽ x ⩽ 0.5$ . Mark uses $f (x)$ as a model to create a barrel. The region enclosed by the graph of $f$ , the $x$ -axis, the line $x = - 0.5$ and the line $x = 0.5$ is rotated 360° about the $x$ -axis. This is shown in the following diagram.

N16/5/MATME/SP2/ENG/TZ0/06

Use the model to find the volume of the barrel.

[3]

The empty barrel is being filled with water. The volume $V m^{3}$ of water in the barrel after $t$ minutes is given by $V = 0.8 (1 - e^{- 0.1 t})$ . How long will it take for the barrel to be half-full?

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to substitute correct limits or the function into the formula involving

$y^{2}$

eg $π \int_{- 0.5}^{0.5} y^{2} d x, π \int {(- 0.8 x^{2} + 0.5)}^{2} d x$

0.601091

volume $= 0.601 (m^{3})$ A2 N3

[3 marks]

attempt to equate half their volume to $V$ (M1)

eg $0.30055 = 0.8 (1 - e^{- 0.1 t})$ , graph

4.71104

4.71 (minutes) A2 N3

[3 marks]

Examiners report

[N/A]

16N.2.SL.TZ0.S_9a

Question

A particle P starts from a point A and moves along a horizontal straight line. Its velocity $v cm s^{- 1}$ after $t$ seconds is given by

$v (t) = {\begin{array}{l} - 2 t + 2, & for 0 ⩽ t ⩽ 1 \\ 3 \sqrt{t} + \frac{4}{t^{2}} - 7, & for 1 ⩽ t ⩽ 12 \end{array}$

The following diagram shows the graph of $v$ .

N16/5/MATME/SP2/ENG/TZ0/09

P is at rest when $t = 1$ and $t = p$ .

When $t = q$ , the acceleration of P is zero.

Find the initial velocity of $P$ .

[2]

Find the value of $p$ .

[2]

(i) Find the value of $q$ .

(ii) Hence, find the speed of P when $t = q$ .

[4]

(i) Find the total distance travelled by P between $t = 1$ and $t = p$ .

(ii) Hence or otherwise, find the displacement of P from A when $t = p$ .

[6]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid attempt to substitute $t = 0$ into the correct function (M1)

eg $- 2 (0) + 2$

2 A1 N2

[2 marks]

recognizing $v = 0$ when P is at rest (M1)

5.21834

$p = 5.22 (seconds)$ A1 N2

[2 marks]

(i) recognizing that $a = v^{'}$ (M1)

eg $v^{'} = 0$ , minimum on graph

1.95343

$q = 1.95$ A1 N2

(ii) valid approach to find their minimum (M1)

eg $v (q), - 1.75879$ , reference to min on graph

1.75879

speed $= 1.76 (c m s^{- 1})$ A1 N2

[4 marks]

(i) substitution of correct $v (t)$ into distance formula, (A1)

eg $\int_{1}^{p} | 3 \sqrt{t} + \frac{4}{t^{2}} - 7 | d t, | \int 3 \sqrt{t} + \frac{4}{t^{2}} - 7 d t |$

4.45368

distance $= 4.45 (cm)$ A1 N2

(ii) displacement from $t = 1$ to $t = p$ (seen anywhere) (A1)

eg $- 4.45368, \int_{1}^{p} (3 \sqrt{t} + \frac{4}{t^{2}} - 7) d t$

displacement from $t = 0$ to $t = 1$ (A1)

eg $\int_{0}^{1} (- 2 t + 2) d t, 0.5 \times 1 \times 2, 1$

valid approach to find displacement for $0 ⩽ t ⩽ p$ M1

eg $\int_{0}^{1} (- 2 t + 2) d t + \int_{1}^{p} (3 \sqrt{t} + \frac{4}{t^{2}} - 7) d t, \int_{0}^{1} (- 2 t + 2) d t - 4.45$

$- 3.45368$

displacement $= - 3.45 (cm)$ A1 N2

[6 marks]

Examiners report

[N/A]

16N.2.SL.TZ0.T_2a

Question

A group of 66 people went on holiday to Hawaii. During their stay, three trips were arranged: a boat trip ( $B$ ), a coach trip ( $C$ ) and a helicopter trip ( $H$ ).

From this group of people:

3	went on all three trips;
16	went on the coach trip only;
13	went on the boat trip only;
5	went on the helicopter trip only;
x	went on the coach trip and the helicopter trip but not the boat trip;
2x	went on the boat trip and the helicopter trip but not the coach trip;
4x	went on the boat trip and the coach trip but not the helicopter trip;
8	did not go on any of the trips.

One person in the group is selected at random.

Draw a Venn diagram to represent the given information, using sets labelled $B$ , $C$ and $H$ .

[5]

Show that $x = 3$ .

[2]

Write down the value of $n (B \cap C)$ .

[1]

Find the probability that this person

(i) went on at most one trip;

(ii) went on the coach trip, given that this person also went on both the helicopter trip and the boat trip.

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

N16/5/MATSD/SP2/ENG/TZ0/02.a/M (A5)

Notes: Award (A1) for rectangle and three labelled intersecting circles (U need not be seen),

(A1) for 3 in the correct region,

(A1) for 8 in the correct region,

(A1) for 5, 13 and 16 in the correct regions,

(A1) for $x$ , $2 x$ and $4 x$ in the correct regions.

[5 marks]

$8 + 13 + 16 + 3 + 5 + x + 2 x + 4 x = 66$ (M1)

Note: Award (M1) for either a completely correct equation or adding all the terms from their diagram in part (a) and equating to 66.

Award (M0)(A0) if their equation has no $x$ .

$7 x = 66 - 45$ OR $7 x + 45 = 66$ (A1)

Note: Award (A1) for adding their like terms correctly, but only when the solution to their equation is equal to 3 and is consistent with their original equation.

$x = 3$ (AG)

Note: The conclusion $x = 3$ must be seen for the (A1) to be awarded.

[2 marks]

15 (A1)(ft)

Note: Follow through from part (a). The answer must be an integer.

[1 mark]

(i) $\frac{42}{66} (\frac{7}{11}, 0.636, 63.6 %)$ (A1)(ft)(A1)(G2)

Note: Award (A1)(ft) for numerator, (A1) for denominator. Follow through from their Venn diagram.

(ii) $\frac{3}{9} (\frac{1}{3}, 0.333, 33.3 %)$ (A1)(A1)(ft)(G2)

Note: Award (A1) for numerator, (A1)(ft) for denominator. Follow through from their Venn diagram.

[4 marks]

Examiners report

[N/A]

16N.2.SL.TZ0.T_6a

Question

A water container is made in the shape of a cylinder with internal height $h$ cm and internal base radius $r$ cm.

N16/5/MATSD/SP2/ENG/TZ0/06

The water container has no top. The inner surfaces of the container are to be coated with a water-resistant material.

The volume of the water container is $0.5 m^{3}$ .

The water container is designed so that the area to be coated is minimized.

One can of water-resistant material coats a surface area of $2000 c m^{2}$ .

Write down a formula for $A$ , the surface area to be coated.

[2]

Express this volume in $c m^{3}$ .

[1]

Write down, in terms of $r$ and $h$ , an equation for the volume of this water container.

[1]

Show that $A = π r^{2} + \frac{1 000 000}{r}$ .

[2]

Find $\frac{d A}{d r}$ .

[3]

Using your answer to part (e), find the value of $r$ which minimizes $A$ .

[3]

Find the value of this minimum area.

[2]

Find the least number of cans of water-resistant material that will coat the area in part (g).

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$(A =) π r^{2} + 2 π r h$ (A1)(A1)

Note: Award (A1) for either $π r^{2}$ OR $2 π r h$ seen. Award (A1) for two correct terms added together.

[2 marks]

$500 000$ (A1)

Notes: Units not required.

[1 mark]

$500 000 = π r^{2} h$ (A1)(ft)

Notes: Award (A1)(ft) for $π r^{2} h$ equating to their part (b).

Do not accept unless $V = π r^{2} h$ is explicitly defined as their part (b).

[1 mark]

$A = π r^{2} + 2 π r (\frac{500 000}{π r^{2}})$ (A1)(ft)(M1)

Note: Award (A1)(ft) for their $\frac{500 000}{π r^{2}}$ seen.

Award (M1) for correctly substituting only $\frac{500 000}{π r^{2}}$ into a correct part (a).

Award (A1)(ft)(M1) for rearranging part (c) to $π r h = \frac{500 000}{r}$ and substituting for $π r h$ in expression for $A$ .

$A = π r^{2} + \frac{1 000 000}{r}$ (AG)

Notes: The conclusion, $A = π r^{2} + \frac{1 000 000}{r}$ , must be consistent with their working seen for the (A1) to be awarded.

Accept $10^{6}$ as equivalent to $1 000 000$ .

[2 marks]

$2 π r - \frac{1 000 000}{r^{2}}$ (A1)(A1)(A1)

Note: Award (A1) for $2 π r$ , (A1) for $\frac{1}{r^{2}}$ or $r^{- 2}$ , (A1) for $- 1 000 000$ .

[3 marks]

$2 π r - \frac{1 000 000}{r^{2}} = 0$ (M1)

Note: Award (M1) for equating their part (e) to zero.

$r^{3} = \frac{1 000 000}{2 π}$ OR $r = \sqrt[3]{\frac{1 000 000}{2 π}}$ (M1)

Note: Award (M1) for isolating $r$ .

sketch of derivative function (M1)

with its zero indicated (M1)

$(r =) 54.2 (cm) (54.1926 \dots)$ (A1)(ft)(G2)

[3 marks]

$π (54.1926 \dots)^{2} + \frac{1 000 000}{(54.1926 \dots)}$ (M1)

Note: Award (M1) for correct substitution of their part (f) into the given equation.

$= 27 700 (c m^{2}) (27 679.0 \dots)$ (A1)(ft)(G2)

[2 marks]

$\frac{27 679.0 \dots}{2000}$ (M1)

Note: Award (M1) for dividing their part (g) by 2000.

$= 13.8395 \dots$ (A1)(ft)

Notes: Follow through from part (g).

14 (cans) (A1)(ft)(G3)

Notes: Final (A1) awarded for rounding up their $13.8395 \dots$ to the next integer.

[3 marks]

Examiners report

[N/A]

17M.2.SL.TZ1.S_6

Question

Let $f (x) = (x^{2} + 3)^{7}$ . Find the term in $x^{5}$ in the expansion of the derivative, $f^{'} (x)$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

derivative of $f (x)$ A2

$7 (x^{2} + 3)^{6} (x 2)$

recognizing need to find $x^{4}$ term in $(x^{2} + 3)^{6}$ (seen anywhere) R1

eg $14 x (termin x^{4})$

valid approach to find the terms in $(x^{2} + 3)^{6}$ (M1)

eg $(\begin{matrix} 6 \\ r \end{matrix}) (x^{2})^{6 - r} (3)^{r}, (x^{2})^{6} (3)^{0} + (x^{2})^{5} (3)^{1} + \dots$ , Pascal’s triangle to 6th row

identifying correct term (may be indicated in expansion) (A1)

eg $5thterm, r = 2, (\begin{matrix} 6 \\ 4 \end{matrix}), (x^{2})^{2} (3)^{4}$

correct working (may be seen in expansion) (A1)

eg $(\begin{matrix} 6 \\ 4 \end{matrix}) (x^{2})^{2} (3)^{4}, 15 \times 3^{4}, 14 x \times 15 \times 81 (x^{2})^{2}$

$17010 x^{5}$ A1 N3

METHOD 2

recognition of need to find $x^{6}$ in $(x^{2} + 3)^{7}$ (seen anywhere) R1

valid approach to find the terms in $(x^{2} + 3)^{7}$ (M1)

eg $(\begin{matrix} 7 \\ r \end{matrix}) (x^{2})^{7 - r} (3)^{r}, (x^{2})^{7} (3)^{0} + (x^{2})^{6} (3)^{1} + \dots$ , Pascal’s triangle to 7th row

identifying correct term (may be indicated in expansion) (A1)

eg6th term, $r = 3, (\begin{matrix} 7 \\ 3 \end{matrix}), (x^{2})^{3} (3)^{4}$

correct working (may be seen in expansion) (A1)

eg $(\begin{matrix} 7 \\ 4 \end{matrix}) (x^{2})^{3} (3)^{4}, 35 \times 3^{4}$

correct term (A1)

$2835 x^{6}$

differentiating their term in $x^{6}$ (M1)

eg $(2835 x^{6})^{'}, (6)(2835 x^{5})$

$17010 x^{5}$ A1 N3

[7 marks]

Examiners report

[N/A]

17M.2.SL.TZ2.S_7

Question

Note: In this question, distance is in metres and time is in seconds.

A particle moves along a horizontal line starting at a fixed point A. The velocity $v$ of the particle, at time $t$ , is given by $v (t) = \frac{2 t^{2} - 4 t}{t^{2} - 2 t + 2}$ , for $0 ⩽ t ⩽ 5$ . The following diagram shows the graph of $v$

M17/5/MATME/SP2/ENG/TZ2/07

There are $t$ -intercepts at $(0, 0)$ and $(2, 0)$ .

Find the maximum distance of the particle from A during the time $0 ⩽ t ⩽ 5$ and justify your answer.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1 (displacement)

recognizing $s = \int v d t$ (M1)

consideration of displacement at $t = 2$ and $t = 5$ (seen anywhere) M1

eg $\int_{0}^{2} v$ and $\int_{0}^{5} v$

Note: Must have both for any further marks.

correct displacement at $t = 2$ and $t = 5$ (seen anywhere) A1A1

$- 2.28318$ (accept 2.28318), 1.55513

valid reasoning comparing correct displacements R1

eg $| - 2.28 | > | 1.56 |$ , more left than right

2.28 (m) A1 N1

Note: Do not award the final A1 without the R1.

METHOD 2 (distance travelled)

recognizing distance $= \int | v | d t$ (M1)

consideration of distance travelled from $t = 0$ to 2 and $t = 2$ to 5 (seen anywhere) M1

eg $\int_{0}^{2} v$ and $\int_{2}^{5} v$

Note: Must have both for any further marks

correct distances travelled (seen anywhere) A1A1

2.28318, (accept $- 2.28318$ ), 3.83832

valid reasoning comparing correct distance values R1

eg $3.84 - 2.28 < 2.28, 3.84 < 2 \times 2.28$

2.28 (m) A1 N1

Note: Do not award the final A1 without the R1.

[6 marks]

Examiners report

[N/A]

17M.2.SL.TZ1.S_7b

Question

A particle P moves along a straight line. Its velocity $v_{P} m s^{- 1}$ after $t$ seconds is given by $v_{P} = \sqrt{t} \sin (\frac{π}{2} t)$ , for $0 ⩽ t ⩽ 8$ . The following diagram shows the graph of $v_{P}$ .

M17/5/MATME/SP2/ENG/TZ1/07

Write down the first value of $t$ at which P changes direction.

[1]

a.i.

Find the total distance travelled by P, for $0 ⩽ t ⩽ 8$ .

[2]

a.ii.

A second particle Q also moves along a straight line. Its velocity, $v_{Q} m s^{- 1}$ after $t$ seconds is given by $v_{Q} = \sqrt{t}$ for $0 ⩽ t ⩽ 8$ . After $k$ seconds Q has travelled the same total distance as P.

Find $k$ .

[4]

Markscheme

$t = 2$ A1 N1

[1 mark]

a.i.

substitution of limits or function into formula or correct sum (A1)

eg $\int_{0}^{8} | v | d t, \int | v_{Q} | d t, \int_{0}^{2} v d t - \int_{2}^{4} v d t + \int_{4}^{6} v d t - \int_{6}^{8} v d t$

9.64782

distance $= 9.65 (metres)$ A1 N2

[2 marks]

a.ii.

correct approach (A1)

eg $s = \int \sqrt{t}, \int_{0}^{k} \sqrt{t} d t, \int_{0}^{k} | v_{Q} | d t$

correct integration (A1)

eg $\int \sqrt{t} = \frac{2}{3} t^{\frac{3}{2}} + c, {[\frac{2}{3} x^{\frac{3}{2}}]}_{0}^{k}, \frac{2}{3} k^{\frac{3}{2}}$

equating their expression to the distance travelled by their P (M1)

eg $\frac{2}{3} k^{\frac{3}{2}} = 9.65, \int_{0}^{k} \sqrt{t} d t = 9.65$

5.93855

5.94 (seconds) A1 N3

[4 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

17M.2.SL.TZ2.S_8a

Question

Let $f (x) = - 0.5 x^{4} + 3 x^{2} + 2 x$ . The following diagram shows part of the graph of $f$ .

M17/5/MATME/SP2/ENG/TZ2/08

There are $x$ -intercepts at $x = 0$ and at $x = p$ . There is a maximum at A where $x = a$ , and a point of inflexion at B where $x = b$ .

Find the value of $p$ .

[2]

Write down the coordinates of A.

[2]

b.i.

Write down the rate of change of $f$ at A.

[1]

b.ii.

Find the coordinates of B.

[4]

c.i.

Find the the rate of change of $f$ at B.

[3]

c.ii.

Let $R$ be the region enclosed by the graph of $f$ , the $x$ -axis, the line $x = b$ and the line $x = a$ . The region $R$ is rotated 360° about the $x$ -axis. Find the volume of the solid formed.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

evidence of valid approach (M1)

eg $f (x) = 0, y = 0$

2.73205

$p = 2.73$ A1 N2

[2 marks]

1.87938, 8.11721

$(1.88, 8.12)$ A2 N2

[2 marks]

b.i.

rate of change is 0 (do not accept decimals) A1 N1

[1 marks]

b.ii.

METHOD 1 (using GDC)

valid approach M1

eg $f^{″} = 0$ , max/min on $f^{'}, x = - 1$

sketch of either $f^{'}$ or $f^{″}$ , with max/min or root (respectively) (A1)

$x = 1$ A1 N1

Substituting their $x$ value into $f$ (M1)

eg $f (1)$

$y = 4.5$ A1 N1

METHOD 2 (analytical)

$f^{″} = - 6 x^{2} + 6$ A1

setting $f^{″} = 0$ (M1)

$x = 1$ A1 N1

substituting their $x$ value into $f$ (M1)

eg $f (1)$

$y = 4.5$ A1 N1

[4 marks]

c.i.

recognizing rate of change is $f^{'}$ (M1)

eg $y^{'}, f^{'} (1)$

rate of change is 6 A1 N2

[3 marks]

c.ii.

attempt to substitute either limits or the function into formula (M1)

involving $f^{2}$ (accept absence of $π$ and/or $d x$ )

eg $π \int {(- 0.5 x^{4} + 3 x^{2} + 2 x)}^{2} d x, \int_{1}^{1.88} f^{2}$

128.890

$volume = 129$ A2 N3

[3 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

17M.2.SL.TZ1.S_10c

Question

Let $f (x) = \ln x$ and $g (x) = 3 + \ln (\frac{x}{2})$ , for $x > 0$ .

The graph of $g$ can be obtained from the graph of $f$ by two transformations:

$\begin{array}{l} ahorizontalstretchofscalefactor q followedby \\ atranslationof (\begin{matrix} h \\ k \end{matrix}) . \end{array}$

Let $h (x) = g (x) \times \cos (0.1 x)$ , for $0 < x < 4$ . The following diagram shows the graph of $h$ and the line $y = x$ .

M17/5/MATME/SP2/ENG/TZ1/10.b.c

The graph of $h$ intersects the graph of $h^{- 1}$ at two points. These points have $x$ coordinates 0.111 and 3.31 correct to three significant figures.

Write down the value of $q$ ;

[1]

a.i.

Write down the value of $h$ ;

[1]

a.ii.

Write down the value of $k$ .

[1]

a.iii.

Find $\int_{0.111}^{3.31} (h (x) - x) d x$ .

[2]

b.i.

Hence, find the area of the region enclosed by the graphs of $h$ and $h^{- 1}$ .

[3]

b.ii.

Let $d$ be the vertical distance from a point on the graph of $h$ to the line $y = x$ . There is a point $P (a, b)$ on the graph of $h$ where $d$ is a maximum.

Find the coordinates of P, where $0.111 < a < 3.31$ .

[7]

Markscheme

$q = 2$ A1 N1

Note: Accept $q = 1$ , $h = 0$ , and $k = 3 - \ln (2)$ , 2.31 as candidate may have rewritten $g (x)$ as equal to $3 + \ln (x) - \ln (2)$ .

[1 mark]

a.i.

$h = 0$ A1 N1

Note: Accept $q = 1$ , $h = 0$ , and $k = 3 - \ln (2)$ , 2.31 as candidate may have rewritten $g (x)$ as equal to $3 + \ln (x) - \ln (2)$ .

[1 mark]

a.ii.

$k = 3$ A1 N1

Note: Accept $q = 1$ , $h = 0$ , and $k = 3 - \ln (2)$ , 2.31 as candidate may have rewritten $g (x)$ as equal to $3 + \ln (x) - \ln (2)$ .

[1 mark]

a.iii.

2.72409

2.72 A2 N2

[2 marks]

b.i.

recognizing area between $y = x$ and $h$ equals 2.72 (M1)

eg M17/5/MATME/SP2/ENG/TZ1/10.b.ii/M

recognizing graphs of $h$ and $h^{- 1}$ are reflections of each other in $y = x$ (M1)

egarea between $y = x$ and $h$ equals between $y = x$ and $h^{- 1}$

$2 \times 2.72 \int_{0.111}^{3.31} (x - h^{- 1} (x)) d x = 2.72$

5.44819

5.45 A1 N3

[??? marks]

b.ii.

valid attempt to find $d$ (M1)

egdifference in $y$ -coordinates, $d = h (x) - x$

correct expression for $d$ (A1)

eg $(\ln \frac{1}{2} x + 3) (\cos 0.1 x) - x$

valid approach to find when $d$ is a maximum (M1)

egmax on sketch of $d$ , attempt to solve $d^{'} = 0$

0.973679

$x = 0.974$ A2 N4

substituting their $x$ value into $h (x)$ (M1)

2.26938

$y = 2.27$ A1 N2

[7 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

a.iii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

17M.2.SL.TZ2.T_5a

Question

Violeta plans to grow flowers in a rectangular plot. She places a fence to mark out the perimeter of the plot and uses 200 metres of fence. The length of the plot is $x$ metres.

M17/5/MATSD/SP2/ENG/TZ2/05

Violeta places the fence so that the area of the plot is maximized.

By selling her flowers, Violeta earns 2 Bulgarian Levs (BGN) per square metre of the plot.

Show that the width of the plot, in metres, is given by $100 - x$ .

[1]

Write down the area of the plot in terms of $x$ .

[1]

Find the value of $x$ that maximizes the area of the plot.

[2]

Show that Violeta earns 5000 BGN from selling the flowers grown on the plot.

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\frac{200 - 2 x}{2}$ (or equivalent) (M1)

$2 x + 2 y = 200$ (or equivalent) (M1)

Note: Award (M1) for a correct expression leading to $100 - x$ (the $100 - x$ does not need to be seen). The 200 must be seen for the (M1) to be awarded. Do not accept $100 - x$ substituted in the perimeter of the rectangle formula.

$100 - x$ (AG)

[1 mark]

$(area =) x (100 - x)$ OR $- x^{2} + 100 x$ (or equivalent) (A1)

[1 mark]

$x = \frac{- 100}{- 2}$ OR $- 2 x + 100 = 0$ ORgraphical method (M1)

Note: Award (M1) for use of axis of symmetry formula or first derivative equated to zero or a sketch graph.

$x = 50$ (A1)(ft)(G2)

Note: Follow through from part (b), provided x is positive and less than 100.

[2 marks]

$50 (100 - 50) \times 2$ (M1)(M1)

Note: Award (M1) for substituting their $x$ into their formula for area (accept “ $50 \times 50$ ” for the substituted formula), and (M1) for multiplying by 2. Award at most (M0)(M1) if their calculation does not lead to 5000 (BGN), although the 5000 (BGN) does not need to be seen explicitly.

Substitution of 50 into area formula may be seen in part (c).

5000 (BGN) (AG)

[2 marks]

Examiners report

[N/A]

17M.2.SL.TZ2.T_6a

Question

Consider the function $f (x) = - x^{4} + a x^{2} + 5$ , where $a$ is a constant. Part of the graph of $y = f (x)$ is shown below.

M17/5/MATSD/SP2/ENG/TZ2/06

It is known that at the point where $x = 2$ the tangent to the graph of $y = f (x)$ is horizontal.

There are two other points on the graph of $y = f (x)$ at which the tangent is horizontal.

Write down the $y$ -intercept of the graph.

[1]

Find $f^{'} (x)$ .

[2]

Show that $a = 8$ .

[2]

c.i.

Find $f (2)$ .

[2]

c.ii.

Write down the $x$ -coordinates of these two points;

[2]

d.i.

Write down the intervals where the gradient of the graph of $y = f (x)$ is positive.

[2]

d.ii.

Write down the range of $f (x)$ .

[2]

Write down the number of possible solutions to the equation $f (x) = 5$ .

[1]

The equation $f (x) = m$ , where $m \in R$ , has four solutions. Find the possible values of $m$ .

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

5 (A1)

Note: Accept an answer of $(0, 5)$ .

[1 mark]

$(f^{'} (x) =) - 4 x^{3} + 2 a x$ (A1)(A1)

Note: Award (A1) for $- 4 x^{3}$ and (A1) for $+ 2 a x$ . Award at most (A1)(A0) if extra terms are seen.

[2 marks]

$- 4 \times 2^{3} + 2 a \times 2 = 0$ (M1)(M1)

Note: Award (M1) for substitution of $x = 2$ into their derivative, (M1) for equating their derivative, written in terms of $a$ , to 0 leading to a correct answer (note, the 8 does not need to be seen).

$a = 8$ (AG)

[2 marks]

c.i.

$(f (2) =) - 2^{4} + 8 \times 2^{2} + 5$ (M1)

Note: Award (M1) for correct substitution of $x = 2$ and $a = 8$ into the formula of the function.

21 (A1)(G2)

[2 marks]

c.ii.

$(x =) - 2, (x =) 0$ (A1)(A1)

Note: Award (A1) for each correct solution. Award at most (A0)(A1)(ft) if answers are given as $(- 2, 21)$ and $(0, 5)$ or $(- 2, 0)$ and $(0, 0)$ .

[2 marks]

d.i.

$x < - 2, 0 < x < 2$ (A1)(ft)(A1)(ft)

Note: Award (A1)(ft) for $x < - 2$ , follow through from part (d)(i) provided their value is negative.

Award (A1)(ft) for $0 < x < 2$ , follow through only from their 0 from part (d)(i); 2 must be the upper limit.

Accept interval notation.

[2 marks]

d.ii.

$y ⩽ 21$ (A1)(ft)(A1)

Notes: Award (A1)(ft) for 21 seen in an interval or an inequality, (A1) for “ $y ⩽$ ”.

Accept interval notation.

Accept $- \infty < y ⩽ 21$ or $f (x) ⩽ 21$ .

Follow through from their answer to part (c)(ii). Award at most (A1)(ft)(A0) if $x$ is seen instead of $y$ . Do not award the second (A1) if a (finite) lower limit is seen.

[2 marks]

3 (solutions) (A1)

[1 mark]

$5 < m < 21$ or equivalent (A1)(ft)(A1)

Note: Award (A1)(ft) for 5 and 21 seen in an interval or an inequality, (A1) for correct strict inequalities. Follow through from their answers to parts (a) and (c)(ii).

Accept interval notation.

[2 marks]

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

17N.2.SL.TZ0.S_5a

Question

Let $f (x) = 6 - \ln (x^{2} + 2)$ , for $x \in R$ . The graph of $f$ passes through the point $(p, 4)$ , where $p > 0$ .

Find the value of $p$ .

[2]

The following diagram shows part of the graph of $f$ .

N17/5/MATME/SP2/ENG/TZ0/05.b

The region enclosed by the graph of $f$ , the $x$ -axis and the lines $x = - p$ and $x = p$ is rotated 360° about the $x$ -axis. Find the volume of the solid formed.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid approach (M1)

eg $f (p) = 4$ , intersection with $y = 4, \pm 2.32$

2.32143

$p = \sqrt{e^{2} - 2}$ (exact), 2.32 A1 N2

[2 marks]

attempt to substitute either their limits or the function into volume formula (must involve $f^{2}$ , accept reversed limits and absence of $π$ and/or $d x$ , but do not accept any other errors) (M1)

eg $\int_{- 2.32}^{2.32} f^{2}, π \int {(6 - \ln (x^{2} + 2))}^{2} d x, 105.675$

331.989

$volume = 332$ A2 N3

[3 marks]

Examiners report

[N/A]

17N.2.SL.TZ0.S_9a

Question

Note: In this question, distance is in metres and time is in seconds.

A particle P moves in a straight line for five seconds. Its acceleration at time $t$ is given by $a = 3 t^{2} - 14 t + 8$ , for $0 ⩽ t ⩽ 5$ .

When $t = 0$ , the velocity of P is $3 m s^{- 1}$ .

Write down the values of $t$ when $a = 0$ .

[2]

Hence or otherwise, find all possible values of $t$ for which the velocity of P is decreasing.

[2]

Find an expression for the velocity of P at time $t$ .

[6]

Find the total distance travelled by P when its velocity is increasing.

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$t = \frac{2}{3} (exact), 0.667, t = 4$ A1A1 N2

[2 marks]

recognizing that $v$ is decreasing when $a$ is negative (M1)

eg $a < 0, 3 t^{2} - 14 t + 8 ⩽ 0$ , sketch of $a$

correct interval A1 N2

eg $\frac{2}{3} < t < 4$

[2 marks]

valid approach (do not accept a definite integral) (M1)

eg $v \int a$

correct integration (accept missing $c$ ) (A1)(A1)(A1)

$t^{3} - 7 t^{2} + 8 t + c$

substituting $t = 0, v = 3$ , (must have $c$ ) (M1)

eg $3 = 0^{3} - 7 (0^{2}) + 8 (0) + c, c = 3$

$v = t^{3} - 7 t^{2} + 8 t + 3$ A1 N6

[6 marks]

recognizing that $v$ increases outside the interval found in part (b) (M1)

eg $0 < t < \frac{2}{3}, 4 < t < 5$ , diagram

one correct substitution into distance formula (A1)

eg $\int_{0}^{\frac{2}{3}} | v |, \int_{4}^{5} | v |, \int_{\frac{2}{3}}^{4} | v |, \int_{0}^{5} | v |$

one correct pair (A1)

eg3.13580 and 11.0833, 20.9906 and 35.2097

14.2191 A1 N2

$d = 14.2 (m)$

[4 marks]

Examiners report

[N/A]

17N.2.SL.TZ0.T_4b

Question

A company performs an experiment on the efficiency of a liquid that is used to detect a nut allergy.

A group of 60 people took part in the experiment. In this group 26 are allergic to nuts. One person from the group is chosen at random.

A second person is chosen from the group.

When the liquid is added to a person’s blood sample, it is expected to turn blue if the person is allergic to nuts and to turn red if the person is not allergic to nuts.

The company claims that the probability that the test result is correct is 98% for people who are allergic to nuts and 95% for people who are not allergic to nuts.

It is known that 6 in every 1000 adults are allergic to nuts.

This information can be represented in a tree diagram.

N17/5/MATSD/SP2/ENG/TZ0/04.c.d.e.f.g

An adult, who was not part of the original group of 60, is chosen at random and tested using this liquid.

The liquid is used in an office to identify employees who might be allergic to nuts. The liquid turned blue for 38 employees.

Find the probability that both people chosen are not allergic to nuts.

[2]

Copy and complete the tree diagram.

[3]

Find the probability that this adult is allergic to nuts and the liquid turns blue.

[2]

Find the probability that the liquid turns blue.

[3]

Find the probability that the tested adult is allergic to nuts given that the liquid turned blue.

[3]

Estimate the number of employees, from this 38, who are allergic to nuts.

[2]

Markscheme

$\frac{34}{60} \times \frac{33}{59}$ (M1)

Note: Award (M1) for their correct product.

$= 0.317 (\frac{187}{590}, 0.316949 \dots, 31.7 %)$ (A1)(ft)(G2)

Note: Follow through from part (a).

[2 marks]

N17/5/MATSD/SP2/ENG/TZ0/04.c/M (A1)(A1)(A1)

Note: Award (A1) for each correct pair of branches.

[3 marks]

$0.006 \times 0.98$ (M1)

Note: Award (M1) for multiplying 0.006 by 0.98.

$= 0.00588 (\frac{147}{25000}, 0.588 %)$ (A1)(G2)

[2 marks]

$0.006 \times 0.98 + 0.994 \times 0.05 (0.00588 + 0.994 \times 0.05)$ (A1)(ft)(M1)

Note: Award (A1)(ft) for their two correct products, (M1) for adding two products.

$= 0.0556 (0.05558, 5.56 %, \frac{2779}{50000})$ (A1)(ft)(G3)

Note: Follow through from parts (c) and (d).

[3 marks]

$\frac{0.006 \times 0.98}{0.05558}$ (M1)(M1)

Note: Award (M1) for their correct numerator, (M1) for their correct denominator.

$= 0.106 (0.105793 \dots, 10.6 %, \frac{42}{397})$ (A1)(ft)(G3)

Note: Follow through from parts (d) and (e).

[3 marks]

$0.105793 \dots \times 38$ (M1)

Note: Award (M1) for multiplying 38 by their answer to part (f).

$= 4.02 (4.02015 \dots)$ (A1)(ft)(G2)

Notes: Follow through from part (f). Use of 3 sf result from part (f) results in an answer of 4.03 (4.028).

[2 marks]

Examiners report

[N/A]

17N.2.SL.TZ0.T_5d

Question

A function $f$ is given by $f (x) = (2 x + 2) (5 - x^{2})$ .

The graph of the function $g (x) = 5^{x} + 6 x - 6$ intersects the graph of $f$ .

Expand the expression for $f (x)$ .

[1]

b.i.

Find $f^{'} (x)$ .

[3]

b.ii.

Draw the graph of $f$ for $- 3 ⩽ x ⩽ 3$ and $- 40 ⩽ y ⩽ 20$ . Use a scale of 2 cm to represent 1 unit on the $x$ -axis and 1 cm to represent 5 units on the $y$ -axis.

[4]

Write down the coordinates of the point of intersection.

[2]

Markscheme

$10 x - 2 x^{3} + 10 - 2 x^{2}$ (A1)

Notes: The expansion may be seen in part (b)(ii).

[1 mark]

b.i.

$10 - 6 x^{2} - 4 x$ (A1)(ft)(A1)(ft)(A1)(ft)

Notes: Follow through from part (b)(i). Award (A1)(ft) for each correct term. Award at most (A1)(ft)(A1)(ft)(A0) if extra terms are seen.

[3 marks]

b.ii.

N17/5/MATSD/SP2/ENG/TZ0/05.d/M (A1)(A1)(ft)(A1)(ft)(A1)

Notes: Award (A1) for correct scale; axes labelled and drawn with a ruler.

Award (A1)(ft) for their correct $x$ -intercepts in approximately correct location.

Award (A1) for correct minimum and maximum points in approximately correct location.

Award (A1) for a smooth continuous curve with approximate correct shape. The curve should be in the given domain.

Follow through from part (a) for the $x$ -intercepts.

[4 marks]

$(1.49, 13.9) ((1.48702 \dots, 13.8714 \dots))$ (G1)(ft)(G1)(ft)

Notes: Award (G1) for 1.49 and (G1) for 13.9 written as a coordinate pair. Award at most (G0)(G1) if parentheses are missing. Accept $x = 1.49$ and $y = 13.9$ . Follow through from part (b)(i).

[2 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

18M.2.SL.TZ1.S_1c

Question

Let f(x) = ln x − 5x , for x > 0 .

Solve f '(x) = f "(x).

Markscheme

METHOD 1 (using GDC)

valid approach (M1)

0.558257

x = 0.558 A1 N2

Note: Do not award A1 if additional answers given.

METHOD 2 (analytical)

attempt to solve their equation f '(x) = f "(x) (do not accept $\frac{1}{x} - 5 = - \frac{1}{x^{2}}$ ) (M1)

eg $5 x^{2} - x - 1 = 0, \frac{1 \pm \sqrt{21}}{10}, \frac{1}{x} = \frac{- 1 \pm \sqrt{21}}{2}, - 0.358$

0.558257

x = 0.558 A1 N2

Note: Do not award A1 if additional answers given.

[2 marks]

Examiners report

[N/A]

18M.2.SL.TZ2.S_3a

Question

Let $f (x) = sin (e^{x})$ for 0 ≤ $x$ ≤ 1.5. The following diagram shows the graph of $f$ .

Find the x-intercept of the graph of $f$ .

[2]

The region enclosed by the graph of $f$ , the y-axis and the x-axis is rotated 360° about the x-axis.

Find the volume of the solid formed.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid approach (M1)
eg $f (x) = 0, e^{x} = 180$ or 0…

1.14472

$x = ln π$ (exact), 1.14 A1 N2

[2 marks]

attempt to substitute either their limits or the function into formula involving $f^{2}$ . (M1)

eg $\int_{0}^{1.14} {f^{2}, π \int (sin (e^{x}))}^{2} d x, 0.795135$

2.49799

volume = 2.50 A2 N3

[3 marks]

Examiners report

[N/A]

18M.2.SL.TZ1.S_4a

Question

Let g(x) = −(x − 1)² + 5.

Let f(x) = x². The following diagram shows part of the graph of f.

The graph of g intersects the graph of f at x = −1 and x = 2.

Write down the coordinates of the vertex of the graph of g.

[1]

On the grid above, sketch the graph of g for −2 ≤ x ≤ 4.

[3]

Find the area of the region enclosed by the graphs of f and g.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

(1,5) (exact) A1 N1

[1 mark]

A1A1A1 N3

Note: The shape must be a concave-down parabola.
Only if the shape is correct, award the following for points in circles:
A1 for vertex,
A1 for correct intersection points,
A1 for correct endpoints.

[3 marks]

integrating and subtracting functions (in any order) (M1)
eg $\int f - g$

correct substitution of limits or functions (accept missing dx, but do not accept any errors, including extra bits) (A1)
eg $\int_{- 1}^{2} g - f, \int - {(x - 1)}^{2} + 5 - x^{2}$

area = 9 (exact) A1 N2

[3 marks]

Examiners report

[N/A]

18M.2.SL.TZ1.S_7a

Question

Let $f (x) = e^{2 sin (\frac{π x}{2})}$ , for x > 0.

The k th maximum point on the graph of f has x-coordinate x_k where $k \in Z^{+}$ .

Given that x_k_{+ 1} = x_k + a, find a.

[4]

Hence find the value of n such that $\sum_{k = 1}^{n} x_{k} = 861$ .

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid approach to find maxima (M1)

eg one correct value of x_k, sketch of f

any two correct consecutive values of x_k (A1)(A1)

eg x₁ = 1, x₂ = 5

a = 4 A1 N3

[4 marks]

recognizing the sequence x_1, x₂_, x_{3, …,} x_n is arithmetic (M1)

eg d = 4

correct expression for sum (A1)

eg $\frac{n}{2} (2 (1) + 4 (n - 1))$

valid attempt to solve for n (M1)

eg graph, 2n² − n − 861 = 0

n = 21 A1 N2

[4 marks]

Examiners report

[N/A]

18M.2.SL.TZ2.S_9a

Question

A particle P moves along a straight line. The velocity v m s⁻¹ of P after t seconds is given by v (t) = 7 cos t − 5t ^{cos t}, for 0 ≤ t ≤ 7.

The following diagram shows the graph of v.

Find the initial velocity of P.

[2]

Find the maximum speed of P.

[3]

Write down the number of times that the acceleration of P is 0 m s⁻² .

[3]

Find the acceleration of P when it changes direction.

[4]

Find the total distance travelled by P.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

initial velocity when t = 0 (M1)

eg v(0)

v = 7 (m s⁻¹) A1 N2

[2 marks]

recognizing maximum speed when $| v |$ is greatest (M1)

eg minimum, maximum, v' = 0

one correct coordinate for minimum (A1)

eg 6.37896, −24.6571

24.7 (ms⁻¹) A1 N2

[3 marks]

recognizing a = v ′ (M1)

eg $a = \frac{d v}{d t}$ , correct derivative of first term

identifying when a = 0 (M1)

eg turning points of v, t-intercepts of v ′

3 A1 N3

[3 marks]

recognizing P changes direction when v = 0 (M1)

t = 0.863851 (A1)

−9.24689

a = −9.25 (ms⁻²) A2 N3

[4 marks]

correct substitution of limits or function into formula (A1)
eg $\int_{0}^{7} | v |, \int_{0}^{0.8638} v d t - \int_{0.8638}^{7} v d t, \int | 7 cos x - 5 x^{cos x} | d x, 3.32 = 60.6$

63.8874

63.9 (metres) A2 N3

[3 marks]

Examiners report

[N/A]

18M.2.SL.TZ1.S_10a

Question

Let $f (x) = 12 cos x - 5 sin x, - π ⩽ x ⩽ 2 π$ , be a periodic function with $f (x) = f (x + 2 π)$

The following diagram shows the graph of $f$ .

There is a maximum point at A. The minimum value of $f$ is −13 .

A ball on a spring is attached to a fixed point O. The ball is then pulled down and released, so that it moves back and forth vertically.

The distance, d centimetres, of the centre of the ball from O at time t seconds, is given by

$d (t) = f (t) + 17, 0 ⩽ t ⩽ 5.$

Find the coordinates of A.

[2]

For the graph of $f$ , write down the amplitude.

[1]

b.i.

For the graph of $f$ , write down the period.

[1]

b.ii.

Hence, write $f (x)$ in the form $p cos (x + r)$ .

[3]

Find the maximum speed of the ball.

[3]

Find the first time when the ball’s speed is changing at a rate of 2 cm s⁻².

[5]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

−0.394791,13

A(−0.395, 13) A1A1 N2

[2 marks]

13 A1 N1

[1 mark]

b.i.

$2 π$ , 6.28 A1 N1

[1 mark]

b.ii.

valid approach (M1)

eg recognizing that amplitude is p or shift is r

$f (x) = 13 cos (x + 0.395)$ (accept p = 13, r = 0.395) A1A1 N3

Note: Accept any value of r of the form $0.395 + 2 π k, k \in Z$

[3 marks]

recognizing need for d ′(t) (M1)

eg −12 sin(t) − 5 cos(t)

correct approach (accept any variable for t) (A1)

eg −13 sin(t + 0.395), sketch of d′, (1.18, −13), t = 4.32

maximum speed = 13 (cms⁻¹) A1 N2

[3 marks]

recognizing that acceleration is needed (M1)

eg a(t), d "(t)

correct equation (accept any variable for t) (A1)

eg $a (t) = - 2, | \frac{d}{d t} (d^{'} (t)) | = 2, - 12 cos (t) + 5 sin (t) = - 2$

valid attempt to solve their equation (M1)

eg sketch, 1.33

1.02154

1.02 A2 N3

[5 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

18M.2.SL.TZ2.T_1d

Question

In a company it is found that 25 % of the employees encountered traffic on their way to work. From those who encountered traffic the probability of being late for work is 80 %.

From those who did not encounter traffic, the probability of being late for work is 15 %.

The tree diagram illustrates the information.

The company investigates the different means of transport used by their employees in the past year to travel to work. It was found that the three most common means of transport used to travel to work were public transportation (P ), car (C ) and bicycle (B ).

The company finds that 20 employees travelled by car, 28 travelled by bicycle and 19 travelled by public transportation in the last year.

Some of the information is shown in the Venn diagram.

There are 54 employees in the company.

Write down the value of a.

[1]

a.i.

Write down the value of b.

[1]

a.ii.

Use the tree diagram to find the probability that an employee encountered traffic and was late for work.

[2]

b.i.

Use the tree diagram to find the probability that an employee was late for work.

[3]

b.ii.

Use the tree diagram to find the probability that an employee encountered traffic given that they were late for work.

[3]

b.iii.

Find the value of x.

[1]

c.i.

Find the value of y.

[1]

c.ii.

Find the number of employees who, in the last year, did not travel to work by car, bicycle or public transportation.

[2]

Find $n ((C \cup B) \cap P^{'})$ .

[2]

Markscheme

a = 0.2 (A1)

[1 mark]

a.i.

b = 0.85 (A1)

[1 mark]

a.ii.

0.25 × 0.8 (M1)

Note: Award (M1) for a correct product.

$= 0.2 (\frac{1}{5}, 20 %)$ (A1)(G2)

[2 marks]

b.i.

0.25 × 0.8 + 0.75 × 0.15 (A1)(ft)(M1)

Note: Award (A1)(ft) for their (0.25 × 0.8) and (0.75 × 0.15), (M1) for adding two products.

$= 0.313 (0.3125, \frac{5}{16}, 31.3 %)$ (A1)(ft)(G3)

Note: Award the final (A1)(ft) only if answer does not exceed 1. Follow through from part (b)(i).

[3 marks]

b.ii.

$\frac{0.25 \times 0.8}{0.25 \times 0.8 + 0.75 \times 0.15}$ (A1)(ft)(A1)(ft)

Note: Award (A1)(ft) for a correct numerator (their part (b)(i)), (A1)(ft) for a correct denominator (their part (b)(ii)). Follow through from parts (b)(i) and (b)(ii).

$= 0.64 (\frac{16}{25}, 64 %)$ (A1)(ft)(G3)

Note: Award final (A1)(ft) only if answer does not exceed 1.

[3 marks]

b.iii.

(x =) 3 (A1)

[1 Mark]

c.i.

(y =) 10 (A1)(ft)

Note: Following through from part (c)(i) but only if their x is less than or equal to 13.

[1 Mark]

c.ii.

54 − (10 + 3 + 4 + 2 + 6 + 8 + 13) (M1)

Note: Award (M1) for subtracting their correct sum from 54. Follow through from their part (c).

= 8 (A1)(ft)(G2)

Note: Award (A1)(ft) only if their sum does not exceed 54. Follow through from their part (c).

[2 marks]

6 + 8 + 13 (M1)

Note: Award (M1) for summing 6, 8 and 13.

27 (A1)(G2)

[2 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

18M.2.SL.TZ1.T_2a

Question

On one day 180 flights arrived at a particular airport. The distance travelled and the arrival status for each incoming flight was recorded. The flight was then classified as on time, slightly delayed, or heavily delayed.

The results are shown in the following table.

A χ² test is carried out at the 10 % significance level to determine whether the arrival status of incoming flights is independent of the distance travelled.

The critical value for this test is 7.779.

A flight is chosen at random from the 180 recorded flights.

State the alternative hypothesis.

[1]

Calculate the expected frequency of flights travelling at most 500 km and arriving slightly delayed.

[2]

Write down the number of degrees of freedom.

[1]

Write down the χ² statistic.

[2]

d.i.

Write down the associated p-value.

[1]

d.ii.

State, with a reason, whether you would reject the null hypothesis.

[2]

Write down the probability that this flight arrived on time.

[2]

Given that this flight was not heavily delayed, find the probability that it travelled between 500 km and 5000 km.

[2]

Two flights are chosen at random from those which were slightly delayed.

Find the probability that each of these flights travelled at least 5000 km.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

The arrival status is dependent on the distance travelled by the incoming flight (A1)

Note: Accept “associated” or “not independent”.

[1 mark]

$\frac{60 \times 45}{180}$ OR $\frac{60}{180} \times \frac{45}{180} \times 180$ (M1)

Note: Award (M1) for correct substitution into expected value formula.

= 15 (A1) (G2)

[2 marks]

4 (A1)

Note: Award (A0) if “2 + 2 = 4” is seen.

[1 mark]

9.55 (9.54671…) (G2)

Note: Award (G1) for an answer of 9.54.

[2 marks]

d.i.

0.0488 (0.0487961…) (G1)

[1 mark]

d.ii.

Reject the Null Hypothesis (A1)(ft)

Note: Follow through from their hypothesis in part (a).

9.55 (9.54671…) > 7.779 (R1)(ft)

0.0488 (0.0487961…) < 0.1 (R1)(ft)

Note: Do not award (A1)(ft)(R0)(ft). Follow through from part (d). Award (R1)(ft) for a correct comparison, (A1)(ft) for a consistent conclusion with the answers to parts (a) and (d). Award (R1)(ft) for χ²_calc > χ²_crit, provided the calculated value is explicitly seen in part (d)(i).

[2 marks]

$\frac{52}{180} (0.289, \frac{13}{45}, 28.9 %)$ (A1)(A1) (G2)

Note: Award (A1) for correct numerator, (A1) for correct denominator.

[2 marks]

$\frac{35}{97} (0.361, 36.1 %)$ (A1)(A1) (G2)

Note: Award (A1) for correct numerator, (A1) for correct denominator.

[2 marks]

$\frac{14}{45} \times \frac{13}{44}$ (A1)(M1)

Note: Award (A1) for two correct fractions and (M1) for multiplying their two fractions.

$= \frac{182}{1980} (0.0919, \frac{91}{990}, 0.091919 \dots, 9.19 %)$ (A1) (G2)

[3 marks]

Examiners report

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

18M.2.SL.TZ1.T_4e

Question

Consider the function $f (x) = \frac{48}{x} + k x^{2} - 58$ , where x > 0 and k is a constant.

The graph of the function passes through the point with coordinates (4 , 2).

P is the minimum point of the graph of f (x).

Sketch the graph of y = f (x) for 0 < x ≤ 6 and −30 ≤ y ≤ 60.
Clearly indicate the minimum point P and the x-intercepts on your graph.

Markscheme

(A1)(A1)(ft)(A1)(ft)(A1)(ft)

Note: Award (A1) for correct window. Axes must be labelled.
(A1)(ft) for a smooth curve with correct shape and zeros in approximately correct positions relative to each other.
(A1)(ft) for point P indicated in approximately the correct position. Follow through from their x-coordinate in part (c). (A1)(ft) for two x-intercepts identified on the graph and curve reflecting asymptotic properties.

[4 marks]

Examiners report

[N/A]

18M.2.SL.TZ1.T_5a

Question

Contestants in a TV gameshow try to get through three walls by passing through doors without falling into a trap. Contestants choose doors at random.
If they avoid a trap they progress to the next wall.
If a contestant falls into a trap they exit the game before the next contestant plays.
Contestants are not allowed to watch each other attempt the game.

The first wall has four doors with a trap behind one door.

Ayako is a contestant.

Natsuko is the second contestant.

The second wall has five doors with a trap behind two of the doors.

The third wall has six doors with a trap behind three of the doors.

The following diagram shows the branches of a probability tree diagram for a contestant in the game.

Write down the probability that Ayako avoids the trap in this wall.

[1]

Find the probability that only one of Ayako and Natsuko falls into a trap while attempting to pass through a door in the first wall.

[3]

Copy the probability tree diagram and write down the relevant probabilities along the branches.

[3]

A contestant is chosen at random. Find the probability that this contestant fell into a trap while attempting to pass through a door in the second wall.

[2]

d.i.

A contestant is chosen at random. Find the probability that this contestant fell into a trap.

[3]

d.ii.

120 contestants attempted this game.

Find the expected number of contestants who fell into a trap while attempting to pass through a door in the third wall.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\frac{3}{4}$ (0.75, 75%) (A1)

[1 mark]

$\frac{3}{4} \times \frac{1}{4} + \frac{1}{4} \times \frac{3}{4}$ OR $2 \times \frac{3}{4} \times \frac{1}{4}$ (M1)(M1)

Note: Award (M1) for their product $\frac{1}{4} \times \frac{3}{4}$ seen, and (M1) for adding their two products or multiplying their product by 2.

$= \frac{3}{8} (\frac{6}{16}, 0.375, 37.5 %)$ (A1)(ft) (G3)

Note: Follow through from part (a), but only if the sum of their two fractions is 1.

[3 marks]

(A1)(ft)(A1)(A1)

Note: Award (A1) for each correct pair of branches. Follow through from part (a).

[3 marks]

$\frac{3}{4} \times \frac{2}{5}$ (M1)

Note: Award (M1) for correct probabilities multiplied together.

$= \frac{3}{10} (0.3, 30 %)$ (A1)(ft) (G2)

Note: Follow through from their tree diagram or part (a).

[2 marks]

d.i.

$1 - \frac{3}{4} \times \frac{2}{5} \times \frac{3}{6}$ OR $\frac{1}{4} + \frac{3}{4} \times \frac{2}{5} + \frac{3}{4} \times \frac{3}{5} \times \frac{3}{6}$ (M1)(M1)

Note: Award (M1) for $\frac{3}{4} \times \frac{3}{5} \times \frac{3}{6}$ and (M1) for subtracting their correct probability from 1, or adding to their $\frac{1}{4} + \frac{3}{4} \times \frac{2}{5}$ .

$= \frac{93}{120} (\frac{31}{40}, 0.775, 77.5 %)$ (A1)(ft) (G2)

Note: Follow through from their tree diagram.

[3 marks]

d.ii.

$\frac{3}{4} \times \frac{3}{5} \times \frac{3}{6} \times 120$ (M1)(M1)

Note: Award (M1) for $\frac{3}{4} \times \frac{3}{5} \times \frac{3}{6} (\frac{3}{4} \times \frac{3}{5} \times \frac{3}{6} OR \frac{27}{120} OR \frac{9}{40})$ and (M1) for multiplying by 120.

= 27 (A1)(ft) (G3)

Note: Follow through from their tree diagram or their $\frac{3}{4} \times \frac{3}{5} \times \frac{3}{6}$ from their calculation in part (d)(ii).

[3 marks]

Examiners report

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

18M.2.SL.TZ2.T_6a

Question

Consider the curve y = 2x³ − 9x² + 12x + 2, for −1 < x < 3

Sketch the curve for −1 < x < 3 and −2 < y < 12.

[4]

A teacher asks her students to make some observations about the curve.

Three students responded.
Nadia said “The x-intercept of the curve is between −1 and zero”.
Rick said “The curve is decreasing when x < 1 ”.
Paula said “The gradient of the curve is less than zero between x = 1 and x = 2 ”.

State the name of the student who made an incorrect observation.

[1]

Find $\frac{dy}{dx}$ .

[3]

Show that the stationary points of the curve are at x = 1 and x = 2.

[2]

Given that y = 2x³ − 9x² + 12x + 2 = k has three solutions, find the possible values of k.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

(A1)(A1)(A1)(A1)

Note: Award (A1) for correct window (condone a window which is slightly off) and axes labels. An indication of window is necessary. −1 to 3 on the x-axis and −2 to 12 on the y-axis and a graph in that window.
(A1) for correct shape (curve having cubic shape and must be smooth).
(A1) for both stationary points in the 1^st quadrant with approximate correct position,
(A1) for intercepts (negative x-intercept and positive y intercept) with approximate correct position.

[4 marks]

Rick (A1)

Note: Award (A0) if extra names stated.

[1 mark]

6x² − 18x + 12 (A1)(A1)(A1)

Note: Award (A1) for each correct term. Award at most (A1)(A1)(A0) if extra terms seen.

[3 marks]

6x² − 18x + 12 = 0 (M1)

Note: Award (M1) for equating their derivative to 0. If the derivative is not explicitly equated to 0, but a subsequent solving of their correct equation is seen, award (M1).

6(x − 1)(x − 2) = 0 (or equivalent) (M1)

Note: Award (M1) for correct factorization. The final (M1) is awarded only if answers are clearly stated.

Award (M0)(M0) for substitution of 1 and of 2 in their derivative.

x = 1, x = 2 (AG)

[2 marks]

6 < k < 7 (A1)(A1)(ft)(A1)

Note: Award (A1) for an inequality with 6, award (A1)(ft) for an inequality with 7 from their part (c) provided it is greater than 6, (A1) for their correct strict inequalities. Accept ]6, 7[ or (6, 7).

[3 marks]

Examiners report

[N/A]

18M.2.SL.TZ1.T_6a

Question

A manufacturer makes trash cans in the form of a cylinder with a hemispherical top. The trash can has a height of 70 cm. The base radius of both the cylinder and the hemispherical top is 20 cm.

A designer is asked to produce a new trash can.

The new trash can will also be in the form of a cylinder with a hemispherical top.

This trash can will have a height of H cm and a base radius of r cm.

There is a design constraint such that H + 2r = 110 cm.

The designer has to maximize the volume of the trash can.

Write down the height of the cylinder.

[1]

Find the total volume of the trash can.

[4]

Find the height of the cylinder, h , of the new trash can, in terms of r.

[2]

Show that the volume, V cm³ , of the new trash can is given by

$V = 110 π r^{3}$ .

[3]

Using your graphic display calculator, find the value of r which maximizes the value of V.

[2]

The designer claims that the new trash can has a capacity that is at least 40% greater than the capacity of the original trash can.

State whether the designer’s claim is correct. Justify your answer.

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

50 (cm) (A1)

[1 mark]

$π \times 50 \times 20^{2} + \frac{1}{2} \times \frac{4}{3} \times π \times 20^{3}$ (M1)(M1)(M1)

Note: Award (M1) for their correctly substituted volume of cylinder, (M1) for correctly substituted volume of sphere formula, (M1) for halving the substituted volume of sphere formula. Award at most (M1)(M1)(M0) if there is no addition of the volumes.

$= 79600 (c m^{3}) (79587.0 \dots (c m^{3}), \frac{76000}{3} π)$ (A1)(ft) (G3)

Note: Follow through from part (a).

[4 marks]

h = H − r (or equivalent) OR H = 110 − 2r (M1)

Note: Award (M1) for writing h in terms of H and r or for writing H in terms of r.

(h =) 110 − 3r (A1) (G2)

[2 marks]

$(V =) \frac{2}{3} π r^{3} + π r^{2} \times (110 - 3 r)$ (M1)(M1)(M1)

Note: Award (M1) for volume of hemisphere, (M1) for correct substitution of their h into the volume of a cylinder, (M1) for addition of two correctly substituted volumes leading to the given answer. Award at most (M1)(M1)(M0) for subsequent working that does not lead to the given answer. Award at most (M1)(M1)(M0) for substituting H = 110 − 2r as their h.

$V = 110 π r^{2} - \frac{7}{3} π r^{3}$ (AG)

[3 marks]

(r =) 31.4 (cm) (31.4285… (cm)) (G2)

$(π) (220 r - 7 r^{2}) = 0$ (M1)

Note: Award (M1) for setting the correct derivative equal to zero.

(r =) 31.4 (cm) (31.4285… (cm)) (A1)

[2 marks]

$(V =) 110 π {(31.4285 \dots)}^{3} - \frac{7}{3} π {(31.4285 \dots)}^{3}$ (M1)

Note: Award (M1) for correct substitution of their 31.4285… into the given equation.

= 114000 (113781…) (A1)(ft)

Note: Follow through from part (e).

(increase in capacity =) $\frac{113.781 \dots - 79587.0 \dots}{79587.0 \dots} \times 100 = 43.0 (%)$ (R1)(ft)

Note: Award (R1)(ft) for finding the correct percentage increase from their two volumes.

1.4 × 79587.0… = 111421.81… (R1)(ft)

Note: Award (R1)(ft) for finding the capacity of a trash can 40% larger than the original.

Claim is correct (A1)(ft)

Note: Follow through from parts (b), (e) and within part (f). The final (R1)(A1)(ft) can be awarded for their correct reason and conclusion. Do not award (R0)(A1)(ft).

[4 marks]

Examiners report

[N/A]

18N.2.SL.TZ0.S_4a

Question

A particle moves along a straight line so that its velocity, $v$ m s⁻¹, after $t$ seconds is given by $v (t) = {1.4}^{t} - 2.7$ , for 0 ≤ $t$ ≤ 5.

Find when the particle is at rest.

[2]

Find the acceleration of the particle when $t = 2$ .

[2]

Find the total distance travelled by the particle.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid approach (M1)

eg $v (t) = 0$ , sketch of graph

2.95195

$t = lo g_{1.4} 2.7$ (exact), $t = 2.95$ (s) A1 N2

[2 marks]

valid approach (M1)

eg $a (t) = v^{'} (t)$ , $v^{'} (2)$

0.659485

$a (2)$ = 1.96 ln 1.4 (exact), $a (2)$ = 0.659 (m s⁻²) A1 N2

[2 marks]

correct approach (A1)

eg $\int_{0}^{5} | v (t) | d t$ , $\int_{0}^{2.95} (- v (t)) d t + \int_{295}^{5} v (t) d t$

5.3479

distance = 5.35 (m) A2 N3

[3 marks]

Examiners report

[N/A]

18N.2.SL.TZ0.S_10a

Question

All lengths in this question are in metres.

Consider the function $f (x) = \sqrt{\frac{4 - x^{2}}{8}}$ , for −2 ≤ $x$ ≤ 2. In the following diagram, the shaded region is enclosed by the graph of $f$ and the $x$ -axis.

A container can be modelled by rotating this region by 360˚ about the $x$ -axis.

Water can flow in and out of the container.

The volume of water in the container is given by the function $g (t)$ , for 0 ≤ $t$ ≤ 4 , where $t$ is measured in hours and $g (t)$ is measured in m³. The rate of change of the volume of water in the container is given by $g^{'} (t) = 0.9 - 2.5 cos (0.4 t^{2})$ .

The volume of water in the container is increasing only when $p$ < $t$ < $q$ .

Find the volume of the container.

[3]

Find the value of $p$ and of $q$ .

[3]

b.i.

During the interval $p$ < $t$ < $q$ , he volume of water in the container increases by $k$ m³. Find the value of $k$ .

[3]

b.ii.

When $t$ = 0, the volume of water in the container is 2.3 m³. It is known that the container is never completely full of water during the 4 hour period.

Find the minimum volume of empty space in the container during the 4 hour period.

[5]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to substitute correct limits or the function into formula involving $f^{2}$ (M1)

eg $π \int_{- 2}^{2} y^{2} d y$ , $π \int {(\sqrt{\frac{4 - x^{2}}{8}})}^{2} d x$

4.18879

volume = 4.19, $\frac{4}{3} π$ (exact) (m³) A2 N3

Note: If candidates have their GDC incorrectly set in degrees, award M marks where appropriate, but no A marks may be awarded. Answers from degrees are p = 13.1243 and q = 26.9768 in (b)(i) and 12.3130 or 28.3505 in (b)(ii).

[3 marks]

recognizing the volume increases when $g^{'}$ is positive (M1)

eg $g^{'} (t)$ > 0, sketch of graph of $g^{'}$ indicating correct interval

1.73387, 3.56393

$p$ = 1.73, $p$ = 3.56 A1A1 N3

[3 marks]

b.i.

valid approach to find change in volume (M1)

eg $g (q) - g (p)$ , $\int_{p}^{q} g^{'} (t) d t$

3.74541

total amount = 3.75 (m³) A2 N3

[3 marks]

b.ii.

Note: There may be slight differences in the final answer, depending on which values candidates carry through from previous parts. Accept answers that are consistent with correct working.

recognizing when the volume of water is a maximum (M1)

eg maximum when $t = q$ , $\int_{0}^{q} g^{'} (t) d t$

valid approach to find maximum volume of water (M1)

eg $2.3 + \int_{0}^{q} g^{'} (t) d t$ , $2.3 + \int_{0}^{p} g^{'} (t) d t + 3.74541$ , 3.85745

correct expression for the difference between volume of container and maximum value (A1)

eg $4.18879 - (2.3 + \int_{0}^{q} g^{'} (t) d t)$ , 4.19 − 3.85745

0.331334

0.331 (m³) A2 N3

[5 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

18N.2.SL.TZ0.T_4a

Question

Consider the function $f (x) = \frac{27}{x^{2}} - 16 x, x \neq 0$ .

Sketch the graph of y = f (x), for −4 ≤ x ≤ 3 and −50 ≤ y ≤ 100.

[4]

Use your graphic display calculator to find the equation of the tangent to the graph of y = f (x) at the point (–2, 38.75).

Give your answer in the form y = mx + c.

[2]

b.iii.

Sketch the graph of the function g (x) = 10x + 40 on the same axes.

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

(A1)(A1)(A1)(A1)

Note: Award (A1) for axis labels and some indication of scale; accept y or f(x).

Use of graph paper is not required. If no scale is given, assume the given window for zero and minimum point.

Award (A1) for smooth curve with correct general shape.

Award (A1) for x-intercept closer to y-axis than to end of sketch.

Award (A1) for correct local minimum with x-coordinate closer to y-axis than end of sketch and y-coordinate less than half way to top of sketch.

Award at most (A1)(A0)(A1)(A1) if the sketch intersects the y-axis or if the sketch curves away from the y-axis as x approaches zero.

[4 marks]

y = −9.25x + 20.3 (y = −9.25x + 20.25) (A1)(A1)

Note: Award (A1) for −9.25x, award (A1) for +20.25, award a maximum of (A0)(A1) if answer is not an equation.

[2 marks]

b.iii.

correct line, y = 10x + 40, seen on sketch (A1)(A1)

Note: Award (A1) for straight line with positive gradient, award (A1) for x-intercept and y-intercept in approximately the correct positions. Award at most (A0)(A1) if ruler not used. If the straight line is drawn on different axes to part (a), award at most (A0)(A1).

[2 marks]

Examiners report

[N/A]

b.iii.

[N/A]

18N.2.SL.TZ0.T_6a

Question

Haruka has an eco-friendly bag in the shape of a cuboid with width 12 cm, length 36 cm and height of 9 cm. The bag is made from five rectangular pieces of cloth and is open at the top.

Nanako decides to make her own eco-friendly bag in the shape of a cuboid such that the surface area is minimized.

The width of Nanako’s bag is x cm, its length is three times its width and its height is y cm.

The volume of Nanako’s bag is 3888 cm³.

Calculate the area of cloth, in cm², needed to make Haruka’s bag.

[2]

Calculate the volume, in cm³, of the bag.

[2]

Use this value to write down, and simplify, the equation in x and y for the volume of Nanako’s bag.

[2]

Write down and simplify an expression in x and y for the area of cloth, A, used to make Nanako’s bag.

[2]

Use your answers to parts (c) and (d) to show that

$A = 3 x^{2} + \frac{10368}{x}$ .

[2]

Find $\frac{d A}{d x}$ .

[3]

Use your answer to part (f) to show that the width of Nanako’s bag is 12 cm.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

36 × 12 + 2(9 ×12) + 2(9 × 36) (M1)

Note: Award (M1) for correct substitution into surface area of cuboid formula.

= 1300 (cm²) (1296 (cm²)) (A1)(G2)

[2 marks]

36 × 9 ×12 (M1)

Note: Award (M1) for correct substitution into volume of cuboid formula.

= 3890 (cm³) (3888 (cm³)) (A1)(G2)

[2 marks]

3 $x$ × $x$ × $y$ = 3888 (M1)

Note: Award (M1) for correct substitution into volume of cuboid formula and equated to 3888.

$x$ ² $y$ = 1296 (A1)(G2)

Note: Award (A1) for correct fully simplified volume of cuboid.

Accept $y = \frac{1296}{x^{2}}$ .

[2 marks]

(A =) 3x² + 2(xy) + 2(3xy) (M1)

Note: Award (M1) for correct substitution into surface area of cuboid formula.

(A =) 3x² + 8xy (A1)(G2)

Note: Award (A1) for correct simplified surface area of cuboid formula.

[2 marks]

$A = 3 x^{2} + 8 x (\frac{1296}{x^{2}})$ (A1)(ft)(M1)

Note: Award (A1)(ft) for correct rearrangement of their part (c) seen (rearrangement may be seen in part(c)), award (M1) for substitution of their part (c) into their part (d) but only if this leads to the given answer, which must be shown.

$A = 3 x^{2} + \frac{10368}{x}$ (AG)

[2 marks]

$(\frac{d A}{d x}) = 6 x - \frac{10368}{x^{2}}$ (A1)(A1)(A1)

Note: Award (A1) for $6 x$ , (A1) for −10368, (A1) for $x^{- 2}$ . Award a maximum of (A1)(A1)(A0) if any extra terms seen.

[3 marks]

$6 x - \frac{10368}{x^{2}} = 0$ (M1)

Note: Award (M1) for equating their $\frac{d A}{d x}$ to zero.

$6 x^{3} = 10368$ OR $6 x^{3} - 10368 = 0$ OR $x^{3} - 1728 = 0$ (M1)

Note: Award (M1) for correctly rearranging their equation so that fractions are removed.

$x = \sqrt[3]{1728}$ (A1)

$x = 12$ (cm) (AG)

Note: The (AG) line must be seen for the final (A1) to be awarded. Substituting $x = 12$ invalidates the method, award a maximum of (M1)(M0)(A0).

[3 marks]

Examiners report

[N/A]

18N.2.SL.TZ0.T_2b.i

Question

160 students attend a dual language school in which the students are taught only in Spanish or taught only in English.

A survey was conducted in order to analyse the number of students studying Biology or Mathematics. The results are shown in the Venn diagram.

Set S represents those students who are taught in Spanish.

Set B represents those students who study Biology.

Set M represents those students who study Mathematics.

A student from the school is chosen at random.

Find the number of students in the school that are taught in Spanish.

[2]

a.i.

Find the number of students in the school that study Mathematics in English.

[2]

a.ii.

Find the number of students in the school that study both Biology and Mathematics.

[2]

a.iii.

Write down $n (S \cap (M \cup B))$ .

[1]

b.i.

Write down $n (B \cap M \cap S^{'})$ .

[1]

b.ii.

Find the probability that this student studies Mathematics.

[2]

c.i.

Find the probability that this student studies neither Biology nor Mathematics.

[2]

c.ii.

Find the probability that this student is taught in Spanish, given that the student studies Biology.

[2]

c.iii.

Markscheme

10 + 40 + 28 + 17 (M1)

= 95 (A1)(G2)

Note: Award (M1) for each correct sum (for example: 10 + 40 + 28 + 17) seen.

[2 marks]

a.i.

20 + 12 (M1)

= 32 (A1)(G2)

Note: Award (M1) for each correct sum (for example: 10 + 40 + 28 + 17) seen.

[2 marks]

a.ii.

12 + 40 (M1)

= 52 (A1)(G2)

Note: Award (M1) for each correct sum (for example: 10 + 40 + 28 + 17) seen.

[2 marks]

a.iii.

78 (A1)

[1 mark]

b.i.

12 (A1)

[1 mark]

b.ii.

$\frac{100}{160}$ $(\frac{5}{8}, 0.625, 62.5 %)$ (A1)(A1) (G2)

Note: Throughout part (c), award (A1) for correct numerator, (A1) for correct denominator. All answers must be probabilities to award (A1).

[2 marks]

c.i.

$\frac{42}{160}$ $(\frac{21}{80}, 0.263 (0.2625), 26.3 % (26.25 %))$ (A1)(A1) (G2)

Note: Throughout part (c), award (A1) for correct numerator, (A1) for correct denominator. All answers must be probabilities to award (A1).

[2 marks]

c.ii.

$\frac{50}{70}$ $(\frac{5}{7}, 0.714 (0.714285 \dots), 71.4 % (71.4285 \dots %))$ (A1)(A1) (G2)

Note: Throughout part (c), award (A1) for correct numerator, (A1) for correct denominator. All answers must be probabilities to award (A1).

[2 marks]

c.iii.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

a.iii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

c.iii.

19M.2.SL.TZ2.S_2a

Question

Let $f (x) = 4 - 2 e^{x}$ . The following diagram shows part of the graph of $f$ .

Find the $x$ -intercept of the graph of $f$ .

[2]

The region enclosed by the graph of $f$ , the $x$ -axis and the $y$ -axis is rotated 360º about the $x$ -axis. Find the volume of the solid formed.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid approach (M1)

eg $f (x) = 0$ , $4 - 2 e^{x} = 0$

0.693147

$x$ = ln 2 (exact), 0.693 A1 N2

[2 marks]

attempt to substitute either their correct limits or the function into formula (M1)

involving $f^{2}$

eg $\int_{0}^{0.693} f^{2}$ , $π \int {(4 - 2 e^{x})}^{2} d x$ , $\int_{0}^{ln 2} {(4 - 2 e^{x})}^{2}$

3.42545

volume = 3.43 A2 N3

[3 marks]

Examiners report

[N/A]

19M.2.SL.TZ1.S_3b

Question

Consider the function $f (x) = x^{2} e^{3 x}$ , $x \in R$ .

The graph of $f$ has a horizontal tangent line at $x = 0$ and at $x = a$ . Find $a$ .

Markscheme

valid method (M1)

eg $f^{'} (x) = 0$ , ,

$a = - 0.667 (= - \frac{2}{3})$ (accept $x = - 0.667$ ) A1 N2

[2 marks]

Examiners report

[N/A]

19M.2.SL.TZ1.S_4a

Question

Let $f^{″} (x) = (cos 2 x) (sin 6 x)$ , for 0 ≤ $x$ ≤ 1.

Sketch the graph of $f^{″}$ on the grid below:

[3]

Find the $x$ -coordinates of the points of inflexion of the graph of $f$ .

[3]

Hence find the values of $x$ for which the graph of $f$ is concave-down.

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

A1A1A1 N3

Note: Only if the shape is approximately correct with exactly 2 maximums and 1 minimum on the interval 0 ≤ $x$ ≤ 0, award the following:
A1 for correct domain with both endpoints within circle and oval.
A1 for passing through the other $x$ -intercepts within the circles.
A1 for passing through the three turning points within circles (ignore $x$ -intercepts and extrema outside of the domain).

[3 marks]

evidence of reasoning (may be seen on graph) (M1)

eg $f^{″} = 0$ , (0.524, 0), (0.785, 0)

0.523598, 0.785398

$x = 0.524 (= \frac{π}{6})$ , $x = 0.785 (= \frac{π}{4})$ A1A1 N3

Note: Award M1A1A0 if any solution outside domain (eg $x = 0$ ) is also included.

[3 marks]

$0.524 < x < 0.785 (\frac{π}{6} < x < \frac{π}{4})$ A2 N2

Note: Award A1 if any correct interval outside domain also included, unless additional solutions already penalized in (b).
Award A0 if any incorrect intervals are also included.

[2 marks]

Examiners report

[N/A]

19M.2.SL.TZ2.S_5a

Question

The population of fish in a lake is modelled by the function

$f (t) = \frac{1000}{1 + 24 e^{- 0.2 t}}$ , 0 ≤ $t$ ≤ 30 , where $t$ is measured in months.

Find the population of fish at $t$ = 10.

[2]

Find the rate at which the population of fish is increasing at $t$ = 10.

[2]

Find the value of $t$ for which the population of fish is increasing most rapidly.

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid approach (M1)

eg $f$ (10)

235.402

235 (fish) (must be an integer) A1 N2

[2 marks]

recognizing rate of change is derivative (M1)

eg rate = $f^{'}$ , $f^{'}$ (10) , sketch of $f^{'}$ , 35 (fish per month)

35.9976

36.0 (fish per month) A1 N2

[2 marks]

valid approach (M1)

eg maximum of $f^{'}$ , $f^{″}$ = 0

15.890

15.9 (months) A1 N2

[2 marks]

Examiners report

[N/A]

19M.2.SL.TZ2.S_8a

Question

In this question distance is in centimetres and time is in seconds.

Particle A is moving along a straight line such that its displacement from a point P, after $t$ seconds, is given by $s_{A} = 15 - t - 6 t^{3} e^{- 0.8 t}$ , 0 ≤ $t$ ≤ 25. This is shown in the following diagram.

Another particle, B, moves along the same line, starting at the same time as particle A. The velocity of particle B is given by $v_{B} = 8 - 2 t$ , 0 ≤ $t$ ≤ 25.

Find the initial displacement of particle A from point P.

[2]

Find the value of $t$ when particle A first reaches point P.

[2]

Find the value of $t$ when particle A first changes direction.

[2]

Find the total distance travelled by particle A in the first 3 seconds.

[3]

Given that particles A and B start at the same point, find the displacement function $s_{B}$ for particle B.

[5]

e.i.

Find the other value of $t$ when particles A and B meet.

[2]

e.ii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid approach (M1)

eg $s_{A} (0), s (0), t = 0$

15 (cm) A1 N2

[2 marks]

valid approach (M1)

eg $s_{A} = 0, s = 0, 6.79321, 14.8651$

2.46941

$t$ = 2.47 (seconds) A1 N2

[2 marks]

recognizing when change in direction occurs (M1)

eg slope of $s$ changes sign, $s^{'} = 0$ , minimum point, 10.0144, (4.08, −4.66)

4.07702

$t$ = 4.08 (seconds) A1 N2

[2 marks]

METHOD 1 (using displacement)

correct displacement or distance from P at $t = 3$ (seen anywhere) (A1)

eg −2.69630, 2.69630

valid approach (M1)

eg 15 + 2.69630, $s (3) - s (0)$ , −17.6963

17.6963

17.7 (cm) A1 N2

METHOD 2 (using velocity)

attempt to substitute either limits or the velocity function into distance formula involving $| v |$ (M1)

eg $\int_{0}^{3} | v | d t$ , $\int | - 1 - 18 t^{2} e^{- 0.8 t} + 4.8 t^{3} e^{- 0.8 t} |$

17.6963

17.7 (cm) A1 N2

[3 marks]

recognize the need to integrate velocity (M1)

eg $\int v (t)$

$8 t - \frac{2 t^{2}}{2} + c$ (accept $x$ instead of $t$ and missing $c$ ) (A2)

substituting initial condition into their integrated expression (must have $c$ ) (M1)

eg $15 = 8 (0) - \frac{2 {(0)}^{2}}{2} + c$ , $c = 15$

$s_{B} (t) = 8 t - t^{2} + 15$ A1 N3

[5 marks]

e.i.

valid approach (M1)

eg $s_{A} = s_{B}$ , sketch, (9.30404, 2.86710)

9.30404

$t = 9.30$ (seconds) A1 N2

Note: If candidates obtain $s_{B} (t) = 8 t - t^{2}$ in part (e)(i), there are 2 solutions for part (e)(ii), 1.32463 and 7.79009. Award the last A1 in part (e)(ii) only if both solutions are given.

[2 marks]

e.ii.

Examiners report

[N/A]

e.i.

[N/A]

e.ii.

19M.2.SL.TZ1.S_9a

Question

Let $f (x) = \frac{16}{x}$ . The line $L$ is tangent to the graph of $f$ at $x = 8$ .

$L$ can be expressed in the form r $= (\begin{matrix} 8 \\ 2 \end{matrix}) + t$ u.

The direction vector of $y = x$ is $(\begin{matrix} 1 \\ 1 \end{matrix})$ .

Find the gradient of $L$ .

[2]

Find u.

[2]

Find the acute angle between $y = x$ and $L$ .

[5]

Find $(f \circ f) (x)$ .

[3]

d.i.

Hence, write down $f^{- 1} (x)$ .

[1]

d.ii.

Hence or otherwise, find the obtuse angle formed by the tangent line to $f$ at $x = 8$ and the tangent line to $f$ at $x = 2$ .

[3]

d.iii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to find $f^{'} (8)$ (M1)

eg $f^{'} (x)$ , $y^{'}$ , $- 16 x^{- 2}$

−0.25 (exact) A1 N2

[2 marks]

u null or any scalar multiple A2 N2

[2 marks]

correct scalar product and magnitudes (A1)(A1)(A1)

scalar product $= 1 \times 4 + 1 \times - 1 (= 3)$

magnitudes $= \sqrt{1^{2} + 1^{2}}$ , $\sqrt{4^{2} + {(- 1)}^{2}}$ $(= \sqrt{2}, \sqrt{17})$

substitution of their values into correct formula (M1)

eg $\frac{4 - 1}{\sqrt{1^{2} + 1^{2}} \sqrt{4^{2} + {(- 1)}^{2}}}$ , $\frac{- 3}{\sqrt{2} \sqrt{17}}$ , 2.1112, 120.96°

1.03037 , 59.0362°

angle = 1.03 , 59.0° A1 N4

[5 marks]

attempt to form composite $(f \circ f) (x)$ (M1)

eg $f (f (x))$ , $f (\frac{16}{x})$ , $\frac{16}{f (x)}$

correct working (A1)

eg $\frac{16}{\frac{16}{x}}$ , $16 \times \frac{x}{16}$

$(f \circ f) (x) = x$ A1 N2

[3 marks]

d.i.

$f^{- 1} (x) = \frac{16}{x}$ (accept $y = \frac{16}{x}$ , $\frac{16}{x}$ ) A1 N1

Note: Award A0 in part (ii) if part (i) is incorrect.
Award A0 in part (ii) if the candidate has found $f^{- 1} (x) = \frac{16}{x}$ by interchanging $x$ and $y$ .

[1 mark]

d.ii.

METHOD 1

recognition of symmetry about $y = x$ (M1)

eg (2, 8) ⇔ (8, 2)

evidence of doubling their angle (M1)

eg $2 \times 1.03$ , $2 \times 59.0$

2.06075, 118.072°

2.06 (radians) (118 degrees) A1 N2

METHOD 2

finding direction vector for tangent line at $x = 2$ (A1)

eg $(\begin{matrix} - 1 \\ 4 \end{matrix})$ , $(\begin{matrix} 1 \\ - 4 \end{matrix})$

substitution of their values into correct formula (must be from vectors) (M1)

eg $\frac{- 4 - 4}{\sqrt{1^{2} + 4^{2}} \sqrt{4^{2} + {(- 1)}^{2}}}$ , $\frac{8}{\sqrt{17} \sqrt{17}}$

2.06075, 118.072°

2.06 (radians) (118 degrees) A1 N2

METHOD 3

using trigonometry to find an angle with the horizontal (M1)

eg $tan θ = - \frac{1}{4}$ , $tan θ = - 4$

finding both angles of rotation (A1)

eg $θ_{1} = 0.244978,14 {.0362}^{\circ}$ , $θ_{1} = 1.81577,104 {.036}^{\circ}$

2.06075, 118.072°

2.06 (radians) (118 degrees) A1 N2

[3 marks]

d.iii.

Examiners report

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

d.iii.

19M.2.SL.TZ2.T_1a

Question

Sila High School has 110 students. They each take exactly one language class from a choice of English, Spanish or Chinese. The following table shows the number of female and male students in the three different language classes.

A $χ^{2}$ test was carried out at the 5 % significance level to analyse the relationship between gender and student choice of language class.

Use your graphic display calculator to write down

The critical value at the 5 % significance level for this test is 5.99.

One student is chosen at random from this school.

Another student is chosen at random from this school.

Write down the null hypothesis, H₀, for this test.

[1]

State the number of degrees of freedom.

[1]

the expected frequency of female students who chose to take the Chinese class.

[1]

c.i.

the $χ^{2}$ statistic.

[2]

c.ii.

State whether or not H₀ should be rejected. Justify your statement.

[2]

Find the probability that the student does not take the Spanish class.

[2]

e.i.

Find the probability that neither of the two students take the Spanish class.

[3]

e.ii.

Find the probability that at least one of the two students is female.

[3]

e.iii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

(H₀:) (choice of) language is independent of gender (A1)

Note: Accept “there is no association between language (choice) and gender”. Accept “language (choice) is not dependent on gender”. Do not accept “not related” or “not correlated” or “not influenced”.

[1 mark]

2 (AG)

[1 mark]

16.4 (16.4181…) (G1)

[1 mark]

c.i.

$χ_{calc}^{2} = 8.69$ (8.68507…) (G2)

[2 marks]

c.ii.

(we) reject the null hypothesis (A1)(ft)

8.68507… > 5.99 (R1)(ft)

Note: Follow through from part (c)(ii). Accept “do not accept” in place of “reject.” Do not award (A1)(ft)(R0).

(we) reject the null hypothesis (A1)

0.0130034 < 0.05 (R1)

Note: Accept “do not accept” in place of “reject.” Do not award (A1)(ft)(R0).

[2 marks]

$\frac{88}{110} (\frac{4}{5}, 0.8, 80 %)$ (A1)(A1)(G2)

Note: Award (A1) for correct numerator, (A1) for correct denominator.

[2 marks]

e.i.

$\frac{88}{110} \times \frac{87}{109}$ (M1)(M1)

Note: Award (M1) for multiplying two fractions. Award (M1) for multiplying their correct fractions.

$(\frac{46}{110}) (\frac{45}{109}) + 2 (\frac{46}{110}) (\frac{42}{109}) + (\frac{42}{110}) (\frac{41}{109})$ (M1)(M1)

Note: Award (M1) for correct products; (M1) for adding 4 products.

$0.639 (0.638532 \dots, \frac{348}{545}, 63.9 %)$ (A1)(ft)(G2)

Note: Follow through from their answer to part (e)(i).

[3 marks]

e.ii.

$1 - \frac{67}{110} \times \frac{66}{109}$ (M1)(M1)

Note: Award (M1) for multiplying two correct fractions. Award (M1) for subtracting their product of two fractions from 1.

$\frac{43}{110} \times \frac{42}{109} + \frac{43}{110} \times \frac{67}{109} + \frac{67}{110} \times \frac{43}{109}$ (M1)(M1)

Note: Award (M1) for correct products; (M1) for adding three products.

$0.631 (0.631192 \dots, 63.1 %, \frac{344}{545})$ (A1)(G2)

[3 marks]

e.iii.

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

e.i.

[N/A]

e.ii.

[N/A]

e.iii.

19M.2.SL.TZ2.T_5b

Question

Consider the function $f (x) = \frac{1}{3} x^{3} + \frac{3}{4} x^{2} - x - 1$ .

The function has one local maximum at $x = p$ and one local minimum at $x = q$ .

Write down the $y$ -intercept of the graph of $y = f (x)$ .

[1]

Sketch the graph of $y = f (x)$ for −3 ≤ $x$ ≤ 3 and −4 ≤ $y$ ≤ 12.

[4]

Determine the range of $f (x)$ for $p$ ≤ $x$ ≤ $q$ .

[3]

Markscheme

−1 (A1)

Note: Accept (0, −1).

[1 mark]

(A1)(A1)(A1)(A1)

Note: Award (A1) for correct window and axes labels, −3 to 3 should be indicated on the $x$ -axis and −4 to 12 on the $y$ -axis.
(A1)) for smooth curve with correct cubic shape;
(A1) for $x$ -intercepts: one close to −3, the second between −1 and 0, and third between 1 and 2; and $y$ -intercept at approximately −1;
(A1) for local minimum in the 4th quadrant and maximum in the 2nd quadrant, in approximately correct positions.
Graph paper does not need to be used. If window not given award at most (A0)(A1)(A0)(A1).

[4 marks]

$- 1.27 ⩽ f (x) ⩽ 1.33 (- 1.27083 \dots ⩽ f (x) ⩽ 1.33333 \dots, - \frac{61}{48} ⩽ f (x) ⩽ \frac{4}{3})$ (A1)(ft)(A1)(ft)(A1)

Note: Award (A1) for −1.27 seen, (A1) for 1.33 seen, and (A1) for correct weak inequalities with their endpoints in the correct order. For example, award (A0)(A0)(A0) for answers like $5 ⩽ f (x) ⩽ 2$ . Accept $y$ in place of $f (x)$ . Accept alternative correct notation such as [−1.27, 1.33].

Follow through from their $p$ and $q$ values from part (g) only if their $f (p)$ and $f (q)$ values are between −4 and 12. Award (A0)(A0)(A0) if their values from (g) are given as the endpoints.

[3 marks]

Examiners report

[N/A]

19N.2.SL.TZ0.S_3a

Question

Let $f (x) = x - 8$ , $g (x) = x^{4} - 3$ and $h (x) = f (g (x))$ .

Find $h (x)$ .

[2]

Let $C$ be a point on the graph of $h$ . The tangent to the graph of $h$ at $C$ is parallel to the graph of $f$ .

Find the $x$ -coordinate of $C$ .

[5]

Markscheme

attempt to form composite (in any order) (M1)

eg $f (x^{4} - 3)$ , ${(x - 8)}^{4} - 3$

$h (x) = x^{4} - 11$ A1 N2

[2 marks]

recognizing that the gradient of the tangent is the derivative (M1)

eg $h^{'}$

correct derivative (seen anywhere) (A1)

$h^{'} (x) = 4 x^{3}$

correct value for gradient of $f$ (seen anywhere) (A1)

$f^{'} (x) = 1$ , $m = 1$

setting their derivative equal to $1$ (M1)

$4 x^{3} = 1$

$0.629960$

$x = \sqrt[3]{\frac{1}{4}}$ (exact), $0.630$ A1 N3

[5 marks]

Examiners report

[N/A]

19N.2.SL.TZ0.S_8a

Question

Let $f (x) = x^{4} - 54 x^{2} + 60 x$ , for $- 1 ⩽ x ⩽ 6$ . The following diagram shows the graph of $f$ .

There are $x$ -intercepts at $x = 0$ and at $x = p$ . There is a maximum at point $A$ where $x = a$ , and a point of inflexion at point $B$ where $x = b$ .

Find the value of $p$ .

[2]

Write down the coordinates of $A$ .

[2]

b.i.

Find the equation of the tangent to the graph of $f$ at $A$ .

[2]

b.ii.

Find the coordinates of $B$ .

[5]

c.i.

Find the rate of change of $f$ at $B$ .

[2]

c.ii.

Let $R$ be the region enclosed by the graph of $f$ , the $x$ -axis and the lines $x = p$ and $x = b$ . The region $R$ is rotated 360º about the $x$ -axis. Find the volume of the solid formed.

[3]

Markscheme

evidence of valid approach (M1)

eg $f (x) = 0$ , $y = 0$

$1.13843$

$p = 1.14$ A1 N2

[2 marks]

$0.562134$ , $16.7641$

$(0.562, 16.8)$ A2 N2

[2 marks]

b.i.

valid approach (M1)

eg tangent at maximum point is horizontal, $f^{'} = 0$

$y = 16.8$ (must be an equation) A1 N2

[2 marks]

b.ii.

METHOD 1 (using GDC)

valid approach M1

eg $f^{″} = 0$ , max/min on $f^{'}$ , $x = - 3$

sketch of either $f^{'}$ or $f^{″}$ , with max/min or root (respectively) (A1)

$x = 3$ A1 N1

substituting their $x$ value into $f$ (M1)

eg $f (3)$

$y = - 225$ (exact) (accept $(3, - 225)$ ) A1 N1

METHOD 2 (analytical)

$f^{″} = 12 x^{2} - 108$ A1

valid approach (M1)

eg $f^{″} = 0$ , $x = \pm 3$

$x = 3$ A1 N1

substituting their $x$ value into $f$ (M1)

eg $f (3)$

$y = - 225$ (exact) (accept $(3, - 225)$ ) A1 N1

[5 marks]

c.i.

recognizing rate of change is $f^{'}$ (M1)

eg $y^{'}$ , $f^{'} (3)$

rate of change is $- 156$ (exact) A1 N2

[2 marks]

c.ii.

attempt to substitute either their limits or the function into volume formula (M1)

eg $\int_{1.14}^{3} f^{2}$ , $π \int {(x^{4} - 54 x^{2} + 60 x)}^{2} d x$ , $25 752.0$

$80 902.3$

volume $= 80 900$ A2 N3

[3 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

19N.2.SL.TZ0.S_10a

Question

A rocket is travelling in a straight line, with an initial velocity of $140$ m s⁻¹. It accelerates to a new velocity of $500$ m s⁻¹ in two stages.

During the first stage its acceleration, $a$ m s⁻², after $t$ seconds is given by $a (t) = 240 sin (2 t)$ , where $0 ⩽ t ⩽ k$ .

The first stage continues for $k$ seconds until the velocity of the rocket reaches $375$ m s⁻¹.

Find an expression for the velocity, $v$ m s⁻¹, of the rocket during the first stage.

[4]

Find the distance that the rocket travels during the first stage.

[4]

During the second stage, the rocket accelerates at a constant rate. The distance which the rocket travels during the second stage is the same as the distance it travels during the first stage.

Find the total time taken for the two stages.

[6]

Markscheme

recognizing that $v = \int a$ (M1)

correct integration A1

eg $- 120 cos (2 t) + c$

attempt to find $c$ using their $v (t)$ (M1)

eg $- 120 cos (0) + c = 140$

$v (t) = - 120 cos (2 t) + 260$ A1 N3

[4 marks]

evidence of valid approach to find time taken in first stage (M1)

eg graph, $- 120 cos (2 t) + 260 = 375$

$k = 1.42595$ A1

attempt to substitute their $v$ and/or their limits into distance formula (M1)

eg $\int_{0}^{1.42595} | v |$ , $\int 260 - 120 cos (2 t)$ , $\int_{0}^{k} (260 - 120 cos (2 t)) d t$

$353.608$

distance is $354$ (m) A1 N3

[4 marks]

recognizing velocity of second stage is linear (seen anywhere) R1

eg graph, $s = \frac{1}{2} h (a + b)$ , $v = m t + c$

valid approach (M1)

eg $\int v = 353.608$

correct equation (A1)

eg $\frac{1}{2} h (375 + 500) = 353.608$

time for stage two $= 0.808248$ ( $0.809142$ from 3 sf) A2

$2.23420$ ( $2.23914$ from 3 sf)

$2.23$ seconds ( $2.24$ from 3 sf) A1 N3

[6 marks]

Examiners report

[N/A]

20N.2.SL.TZ0.S_10a

Question

Consider a function $f (x)$ , for $x \geq 0$ . The derivative of $f$ is given by $f' (x) = \frac{6 x}{x^{2} + 4}$ .

The graph of $f$ is concave-down when $x > n$ .

Show that $f'' (x) = \frac{24 - 6 x^{2}}{{(x^{2} + 4)}^{2}}$ .

[4]

Find the least value of $n$ .

[2]

Find $\int \frac{6 x}{x^{2} + 4} d x$ .

[3]

Let $R$ be the region enclosed by the graph of $f$ , the $x$ -axis and the lines $x = 1$ and $x = 3$ . The area of $R$ is $19.6$ , correct to three significant figures.

Find $f (x)$ .

[7]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

evidence of choosing the quotient rule (M1)

eg $\frac{v u' - u v'}{v^{2}}$

derivative of $6 x$ is $6$ (must be seen in rule) (A1)

derivative of $x^{2} + 4$ is $2 x$ (must be seen in rule) (A1)

correct substitution into the quotient rule A1

eg $\frac{6 (x^{2} + 4) - (6 x) (2 x)}{{(x^{2} + 4)}^{2}}$

$f'' (x) = \frac{24 - 6 x^{2}}{{(x^{2} + 4)}^{2}}$ AG N0

METHOD 2

evidence of choosing the product rule (M1)

eg $v u' + u v'$

derivative of $6 x$ is $6$ (must be seen in rule) (A1)

derivative of ${(x^{2} + 4)}^{- 1}$ is $- 2 x {(x^{2} + 4)}^{- 2}$ (must be seen in rule) (A1)

correct substitution into the product rule A1

eg $6 {(x^{2} + 4)}^{- 1} + (- 1) (6 x) (2 x) {(x^{2} + 4)}^{- 2}$

$f'' (x) = \frac{24 - 6 x^{2}}{{(x^{2} + 4)}^{2}}$ AG N0

[4 marks]

METHOD 1 (2^nd derivative) (M1)

valid approach

eg $f'' < 0, 24 - 6 x^{2} < 0, n = \pm 2, x = 2$

$n = 2$ (exact) A1 N2

METHOD 2 (1^st derivative)

valid attempt to find local maximum on $f'$ (M1)

eg sketch with max indicated, $(2, 1.5), x = 2$

$n = 2$ (exact) A1 N2

[2 marks]

evidence of valid approach using substitution or inspection (M1)

eg $\int 3 (2 x) \frac{1}{u} d x, u = x^{2} + 4, d u = 2 x d x, \int 3 \times (\frac{1}{u}) d u$

$\int \frac{6 x}{(x^{2} + 4)} d x = 3 \ln (x^{2} + 4) + c$ A2 N3

[3 marks]

recognizing that area $= \int_{1}^{3} f (x) d x$ (seen anywhere) (M1)

recognizing that their answer to (c) is their $f (x)$ (accept absence of $c$ ) (M1)

eg $f (x) = 3 \ln (x^{2} + 4) + c, f (x) = 3 \ln (x^{2} + 4)$

correct value for $\int_{1}^{3} 3 \ln (x^{2} + 4) d x$ (seen anywhere) (A1)

eg $12.4859$

correct integration for $\int_{1}^{3} c d x$ (seen anywhere) (A1)

${[c x]}_{1}^{3}, 2 c$

adding their integrated expressions and equating to $19.6$ (do not accept an expression which involves an integral) (M1)

eg $12.4859 + 2 c = 19.6, 2 c = 7.114$

$c = 3.55700$ (A1)

$f (x) = 3 \ln (x^{2} + 4) + 3.56$ A1 N4

[7 marks]

Examiners report

[N/A]

20N.2.SL.TZ0.T_4a

Question

Hyungmin designs a concrete bird bath. The bird bath is supported by a pedestal. This is shown in the diagram.

The interior of the bird bath is in the shape of a cone with radius $r$ , height $h$ and a constant slant height of $50 cm$ .

Let $V$ be the volume of the bird bath.

Hyungmin wants the bird bath to have maximum volume.

Write down an equation in $r$ and $h$ that shows this information.

[1]

Show that $V = \frac{2500 π h}{3} - \frac{π h^{3}}{3}$ .

[1]

Find $\frac{d V}{d h}$ .

[2]

Using your answer to part (c), find the value of $h$ for which $V$ is a maximum.

[2]

Find the maximum volume of the bird bath.

[2]

To prevent leaks, a sealant is applied to the interior surface of the bird bath.

Find the surface area to be covered by the sealant, given that the bird bath has maximum volume.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$h^{2} + r^{2} = 50^{2}$ (or equivalent) (A1)

Note: Accept equivalent expressions such as $r = \sqrt{2500 - h^{2}}$ or $h = \sqrt{2500 - r^{2}}$ . Award (A0) for a final answer of $\pm \sqrt{2500 - h^{2}}$ or $\pm \sqrt{2500 - r^{2}}$ , or any further incorrect working.

[1 mark]

$\frac{1}{3} \times π \times (2500 - h^{2}) \times h$ OR $\frac{1}{3} \times π \times {(\sqrt{2500 - h^{2}})}^{2} \times h$ (M1)

Note: Award (M1) for correct substitution in the volume of cone formula.

$V = \frac{2500 π h}{3} - \frac{π h^{3}}{3}$ (AG)

Note: The final line must be seen, with no incorrect working, for the (M1) to be awarded.

[1 mark]

$(\frac{d V}{d h} =) \frac{2500 π}{3} - π h^{2}$ (A1)(A1)

Note: Award (A1) for $\frac{2500 π}{3}$ , (A1) for $- π h^{2}$ . Award at most (A1)(A0) if extra terms are seen. Award (A0) for the term $- \frac{3 π h^{2}}{3}$ .

[2 marks]

$0 = \frac{2500 π}{3} - π h^{2}$ (M1)

Note: Award (M1) for equating their derivative to zero. Follow through from part (c).

sketch of $\frac{d V}{d h}$ (M1)

Note: Award (M1) for a labelled sketch of $\frac{d V}{d h}$ with the curve/axes correctly labelled or the $x$ -intercept explicitly indicated.

$(h =) 28.9 (cm) (\sqrt{\frac{2500}{3}}, \frac{50}{\sqrt{3}}, \frac{50 \sqrt{3}}{3}, 28.8675 \dots)$ (A1)(ft)

Note: An unsupported $28.9 cm$ is awarded no marks. Graphing the function $V (h)$ is not an acceptable method and (M0)(A0) should be awarded. Follow through from part (c). Given the restraints of the question, $h \geq 50$ is not possible.

[2 marks]

$(V =) \frac{2500 \times π \times 28.8675 \dots}{3} - \frac{π {(28.8675 \dots)}^{3}}{3}$ (M1)

$\frac{1}{3} π {(40.828 \dots)}^{2} \times 28.8675 \dots$ (M1)

Note: Award (M1) for substituting their $28.8675 \dots$ in the volume formula.

$(V =) 50400 ({cm}^{3}) (50383.3 \dots)$ (A1)(ft)(G2)

Note: Follow through from part (d).

[2 marks]

$(S =) π \times \sqrt{2500 - {(28.8675 \dots)}^{2}} \times 50$ (A1)(ft)(M1)

Note: Award (A1) for their correct radius seen $(40.8248 \dots, \sqrt{2500 - {(28.8675 \dots)}^{2}})$ .
Award (M1) for correctly substituted curved surface area formula for a cone.

$(S =) 6410 ({cm}^{2}) (6412.74 \dots)$ (A1)(ft)(G2)

Note: Follow through from parts (a) and (d).

[3 marks]

Examiners report

[N/A]

21M.2.SL.TZ1.1a

Question

Find $\int (6 x + 7) d x$ .

[3]

Given $f' (x) = 6 x + 7$ and $f (1.2) = 7.32$ , find $f (x)$ .

[3]

Markscheme

correct integration $3 x^{2} + 7 x + c$ A1A1A1

Note: Award A1 for $3 x^{2}$ , A1 for $7 x$ and A1 for $+ c$

[3 marks]

recognition that $f (x) = \int f' (x) d x$ (M1)

$3 {(1.2)}^{2} + 7 (1.2) + c = 7.32$ (A1)

$c = - 5.4$

$f (x) = 3 x^{2} + 7 x - 5.4$ A1

[3 marks]

Examiners report

[N/A]

21M.2.SL.TZ1.5a

Question

A particle moves in a straight line. The velocity, $v {ms}^{- 1}$ , of the particle at time $t$ seconds is given by $v (t) = t \sin t - 3$ , for $0 \leq t \leq 10$ .

The following diagram shows the graph of $v$ .

Find the smallest value of $t$ for which the particle is at rest.

[2]

Find the total distance travelled by the particle.

[2]

Find the acceleration of the particle when $t = 7$ .

[2]

Markscheme

recognising $v = 0$ (M1)

$t = 6.74416 \dots$

$= 6.74$ (sec) A1

Note: Do not award A1 if additional values are given.

[2 marks]

$\int_{0}^{10} |v (t)| d t$ OR $- \int_{0}^{6.74416 \dots} v (t) d t + \int_{6.74416 \dots}^{9.08837 \dots} v (t) d t - \int_{9.08837 \dots}^{10} v (t) d t$ (A1)

$= 37.0968 \dots$

$= 37.1 (m)$ A1

[2 marks]

recognizing acceleration at $t = 7$ is given by $v' (7)$ (M1)

acceleration $= 5.93430 \dots$

$= 5.93 ({ms}^{- 2})$ A1

[2 marks]

Examiners report

[N/A]

21M.2.SL.TZ1.9b

Question

Consider the function $f$ defined by $f (x) = 90 e^{- 0.5 x}$ for $x \in ℝ^{+}$ .

The graph of $f$ and the line $y = x$ intersect at point $P$ .

The line $L$ has a gradient of $- 1$ and is a tangent to the graph of $f$ at the point $Q$ .

The shaded region $A$ is enclosed by the graph of $f$ and the lines $y = x$ and $L$ .

Find the $x$ -coordinate of $P$ .

[2]

Find the exact coordinates of $Q$ .

[4]

Show that the equation of $L$ is $y = - x + 2 \ln 45 + 2$ .

[2]

Find the $x$ -coordinate of the point where $L$ intersects the line $y = x$ .

[1]

d.i.

Hence, find the area of $A$ .

[4]

d.ii.

The line $L$ is tangent to the graphs of both $f$ and the inverse function $f^{- 1}$ .

Find the shaded area enclosed by the graphs of $f$ and $f^{- 1}$ and the line $L$ .

[2]

Markscheme

Attempt to find the point of intersection of the graph of $f$ and the line $y = x$ (M1)

$x = 5.56619 \dots$

$= 5.57$ A1

[2 marks]

$f' (x) = - 45 e^{- 0.5 x}$ A1

attempt to set the gradient of $f$ equal to $- 1$ (M1)

$- 45 e^{- 0.5 x} = -1$

$Q$ has coordinates $(2 \ln 45, 2)$ (accept ( $- 2 \ln \frac{1}{45}, 2$ ) A1A1

Note: Award A1 for each value, even if the answer is not given as a coordinate pair.

Do not accept $\frac{\ln \frac{1}{45}}{- 0.5}$ or $\frac{\ln 45}{0.5}$ as a final value for $x$ . Do not accept $2.0$ or $2.00$ as a final value for $y$ .

[4 marks]

attempt to substitute coordinates of $Q$ (in any order) into an appropriate equation (M1)

$y - 2 = - (x - 2 \ln 45)$ OR $2 = - 2 \ln 45 + c$ A1

equation of $L$ is $y = - x + 2 \ln 45 + 2$ AG

[2 marks]

$x = \ln 45 + 1 (= 4.81)$ A1

[1 mark]

d.i.

appropriate method to find the sum of two areas using integrals of the difference of two functions (M1)

Note: Allow absence of incorrect limits.

$\int_{4.806 \dots}^{5.566 \dots} (x - (- x + 2 \ln 45 + 2)) d x + \int_{5.566 \dots}^{7.613 \dots} (90 e^{- 0.5 x} - (- x + 2 \ln 45 + 2)) d x$ (A1)(A1)

Note: Award A1 for one correct integral expression including correct limits and integrand.
Award A1 for a second correct integral expression including correct limits and integrand.

$= 1.52196 \dots$

$= 1.52$ A1

[4 marks]

d.ii.

by symmetry $2 \times 1.52 \dots$ (M1)

$= 3.04$ A1

Note: Accept any answer that rounds to $3.0$ (but do not accept $3$ ).

[2 marks]

Examiners report

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

22M.2.SL.TZ2.2

Question

The derivative of a function $g$ is given by $g' (x) = 3 x^{2} + 5 e^{x}$ , where $x \in ℝ$ . The graph of $g$ passes through the point $(0, 4)$ . Find $g (x)$ .

Markscheme

METHOD 1

recognises that $g (x) = \int (3 x^{2} + 5 e^{x}) d x$ (M1)

$g (x) = x^{3} + 5 e^{x} (+ C)$ (A1)(A1)

Note: Award A1 for each integrated term.

substitutes $x = 0$ and $y = 4$ into their integrated function (must involve $+ C$ ) (M1)

$4 = 0 + 5 + C \Rightarrow C = - 1$

$g (x) = x^{3} + 5 e^{x} - 1$ A1

METHOD 2

attempts to write both sides in the form of a definite integral (M1)

$\int_{0}^{x} g' (t) d t = \int_{0}^{x} (3 t^{2} + 5 e^{t}) d t$ (A1)

$g (x) - 4 = x^{3} + 5 e^{x} - 5 e^{0}$ (A1)(A1)

Note: Award A1 for $g (x) - 4$ and A1 for $x^{3} + 5 e^{x} - 5 e^{0}$ .

$g (x) = x^{3} + 5 e^{x} - 1$ A1

[5 marks]

Examiners report

While many students were successful in solving this question, some did not consider the constant of integration or struggled to integrate the exponential term. A few students lost the final mark for stopping at $C = - 1$ and not giving the formula for $g (x)$ .

22M.2.SL.TZ1.5a

Question

A particle moves along a straight line so that its velocity, $v {m s}^{- 1}$ , after $t$ seconds is given by $v (t) = e^{\sin t} + 4 \sin t$ for $0 \leq t \leq 6$ .

Find the value of $t$ when the particle is at rest.

[2]

Find the acceleration of the particle when it changes direction.

[3]

Find the total distance travelled by the particle.

[2]

Markscheme

recognizing at rest $v = 0$ (M1)

$t = 3.34692 \dots$

$t = 3.35$ (seconds) A1

Note: Award (M1)A0 for additional solutions to $v = 0$ eg $t = - 0.205$ or $t = 6.08$ .

[2 marks]

recognizing particle changes direction when $v = 0$ OR when $t = 3.34692 \dots$ (M1)

$a = - 4.71439 \dots$

$a = - 4.71 ({ms}^{- 2})$ A2

[3 marks]

distance travelled $= \int_{0}^{6} |v| d t$ OR

$\int_{0}^{3.34 \dots} (e^{\sin (t)} + 4 \sin (t)) d t - \int_{3.34 \dots}^{6} (e^{\sin (t)} + 4 \sin (t)) d t (= 14.3104 \dots + 6.44300 \dots)$ (A1)

$= 20.7534 \dots$

$= 20.8$ (metres) A1

[2 marks]

Examiners report

The majority of candidates found this question challenging but were often able to gain some of the marks in each part. However, it was not uncommon to see candidates manage either all of this question, or none of it, which unfortunately suggested that not all candidates had covered this content.

In part (a), while many recognized $v = 0$ when the particle is at rest, a common error was to assume that $t = 0$ . It was pleasing to see the majority of those who had the correct equation, manage to progress to the correct value of t, having recognised the domain and the angle measure. Inevitably, a few candidates ignored the domain and obtained $t = - 0.205$ , or found $t = 0$ .

Part (b) was not well done. The most successful approach was to use the GDC to find the gradient of the curve at the value of t obtained in part (a). This was well communicated, concise and generally accurate, although some either rounded incorrectly, or obtained $a = 4.71$ rather than $a = - 4.71$ . Of those that did not use the GDC, many were aware that $a (t) = v' (t)$ and made an attempt to find an expression for $v' (t)$ . However, the majority appeared not to recognise when the particle would change direction and went on to substitute an incorrect value of $t$ , not realising that they had obtained the required value earlier.

Those that had been successful in parts (a) and (b), particularly if they had been using their GDC, were generally able to complete part (c). However, it was disappointing that many candidates who understood that an integral was required, did not refer to the formula booklet and omitted the absolute value from the integrand.

[N/A]

22M.2.SL.TZ2.6a

Question

A particle moves in a straight line such that its velocity, $v m s^{- 1}$ , at time $t$ seconds is given by $v = \frac{(t^{2} + 1) \cos t}{4}, 0 \leq t \leq 3$ .

Determine when the particle changes its direction of motion.

[2]

Find the times when the particle’s acceleration is $- 1.9 {m s}^{- 2}$ .

[3]

Find the particle’s acceleration when its speed is at its greatest.

[2]

Markscheme

recognises the need to find the value of $t$ when $v = 0$ (M1)

$t = 1.57079 \dots (= \frac{π}{2})$

$t = 1.57 (= \frac{π}{2})$ (s) A1

[2 marks]

recognises that $a (t) = v' (t)$ (M1)

$t_{1} = 2.26277 \dots, t_{2} = 2.95736 \dots$

$t_{1} = 2.26, t_{2} = 2.96$ (s) A1A1

Note: Award M1A1A0 if the two correct answers are given with additional values outside $0 \leq t \leq 3$ .

[3 marks]

speed is greatest at $t = 3$ (A1)

$a = - 1.83778 \dots$

$a = - 1.84 ({m s}^{- 2})$ A1

[2 marks]

Examiners report

In part (a) many did not realize the change of motion occurred when $v = 0$ . A common error was finding $v (0)$ or thinking that it was at the maximum of $v$ .

In part (b), most candidates knew to differentiate but some tried to substitute in $- 1.9$ for $t$ , while others struggled to differentiate the function by hand rather than using the GDC. Many candidates tried to solve the equation analytically and did not use their technology. Of those who did, many had their calculators in degree mode.

Almost all candidates who attempted part (c) thought the greatest speed was the same as the maximum of $v$ .

[N/A]

22M.2.SL.TZ1.7d

Question

All lengths in this question are in centimetres.

A solid metal ornament is in the shape of a right pyramid, with vertex $V$ and square base $ABCD$ . The centre of the base is $X$ . Point $V$ has coordinates $(1, 5, 0)$ and point $A$ has coordinates $(- 1, 1, 6)$ .

The volume of the pyramid is $57.2 {cm}^{3}$ , correct to three significant figures.

Find $AV$ .

[2]

Given that $A \hat{V} B = 40 °$ , find $AB$ .

[3]

Find the height of the pyramid, $VX$ .

[3]

A second ornament is in the shape of a cuboid with a rectangular base of length $2 x cm$ , width $x cm$ and height $y cm$ . The cuboid has the same volume as the pyramid.

The cuboid has a minimum surface area of $S {cm}^{2}$ . Find the value of $S$ .

[5]

Markscheme

attempt to use the distance formula to find $AV$ (M1)

$\sqrt{{(1 - (- 1))}^{2} + {(5 - 1)}^{2} + {(0 - 6)}^{2}}$

$= 7.48331 \dots$

$= 7.48 (cm) (= \sqrt{56} or 2 \sqrt{14})$ A1

[2 marks]

METHOD 1

attempt to apply cosine rule OR sine rule to find $AB$ (M1)

$(AB =) \sqrt{7.48 \dots^{2} + 7.48 \dots^{2} - 2 \times 7.48 \dots \times 7.48 \dots \cos (40 °)}$ OR $\frac{AB}{\sin 40 °} = \frac{\sqrt{56}}{\sin 70 °}$ (A1)

$= 5.11888 \dots$

$= 5.12 (cm)$ A1

METHOD 2

Let $M$ be the midpoint of $[AB]$

attempt to apply right-angled trigonometry on triangle $AVM$ (M1)

$= 2 \times 7.48 \dots \times \sin (20 °)$ (A1)

$= 5.11888 \dots$

$= 5.12 (cm)$ A1

[3 marks]

METHOD 1

equating volume of pyramid formula to $57.2$ (M1)

$\frac{1}{3} \times 5.11 \dots^{2} \times h = 57.2$ (A1)

$h = 6.54886 \dots$

$h = 6.55 (cm)$ A1

METHOD 2

Let $M$ be the midpoint of $[AB]$

${AV}^{2} = {AM}^{2} + {MX}^{2} + {XV}^{2}$ (M1)

$\Rightarrow XV = \sqrt{7.48 \dots^{2} - {(\frac{5.11 \dots}{2})}^{2} - {(\frac{5.11 \dots}{2})}^{2}}$ (A1)

$h = 6.54886 \dots$

$h = 6.55 (cm)$ A1

[3 marks]

$V = x \times 2 x \times y = 57.2$ (A1)

$S = 2 (2 x^{2} + x y + 2 x y)$ A1

Note: Condone use of $A$ .

attempt to substitute $y = \frac{57.2}{2 x^{2}}$ into their expression for surface area (M1)

$(S (x) =) 4 x^{2} + 6 x (\frac{57.2}{2 x^{2}})$

EITHER

attempt to find minimum turning point on graph of area function (M1)

$\frac{d S}{d x} = 8 x - 171.6 x^{- 2} = 0$ OR $x = 2.77849 \dots$ (M1)

THEN

$92.6401 \dots$

minimum surface area $= 92.6 ({cm}^{2})$ A1

[5 marks]

Examiners report

Parts (a), (b) and (c) were completed well by many candidates, but few were able to make any significant progress in part (d).

In part (a), many candidates were able to apply the distance formula and successfully find AV. However, a common error was to work in two-dimensions and to apply Pythagoras' Theorem once, neglecting completely the z-coordinates. Many recognised the need to use either the sine or cosine rule in part (b) to find the length AB. Common errors in this part included: the GDC being set incorrectly in radians; applying right-angled trigonometry on a 40°, 40°, 70° triangle; or using $\frac{1}{2} AB$ as the length of AX in triangle AVX.

Despite the formula for the volume of a pyramid being in the formula booklet, a common error in part (c) was to omit the factor of $\frac{1}{3}$ from the volume formula, or not to recognise that the area of the base of the pyramid was ${AB}^{2}$ .

The most challenging part of this question proved to be the optimization of the surface area of the cuboid in part (d). Although some candidates were able to form an equation involving the volume of the cuboid, an expression for the surface area eluded most. A common error was to gain a surface area which involved eight sides rather than six. It was surprising that few who were able to find both the equation and an expression were able to progress any further. Of those that did, few used their GDC to find the minimum surface area directly, with most preferring the more time consuming analytical approach.

[N/A]

22M.2.SL.TZ2.8c

Question

The scientist found that the height of Plant $A, h_{A} cm$ , at time $t$ weeks can be modelled by the function $h_{A} (t) = \sin (2 t + 6) + 9 t + 27$ , where $0 \leq t \leq 9$ .

The scientist found that the height of Plant $B, h_{B} cm$ , at time $t$ weeks can be modelled by the function $h_{B} (t) = 8 t + 32$ , where $0 \leq t \leq 9$ .

Use the scientist’s models to find the initial height of

Plant $B$ .

[1]

a.i.

Plant $A$ correct to three significant figures.

[2]

a.ii.

Find the values of $t$ when $h_{A} (t) = h_{B} (t)$ .

[3]

For $0 \leq t \leq 9$ , find the total amount of time when the rate of growth of Plant $B$ was greater than the rate of growth of Plant $A$ .

[6]

Markscheme

$32$ (cm) A1

[1 mark]

a.i.

$h_{A} (0) = \sin (6) + 27$ (M1)

$= 26.7205 \dots$

$= 26.7$ (cm) A1

[2 marks]

a.ii.

attempts to solve $h_{A} (t) = h_{B} (t)$ for $t$ (M1)

$t = 4.00746 \dots, 4.70343 \dots, 5.88332 \dots$

$t = 4.01, 4.70, 5.88$ (weeks) A2

[3 marks]

recognises that $h_{A}' (t)$ and $h_{B}' (t)$ are required (M1)

attempts to solve $h_{A}' (t) = h_{B}' (t)$ for $t$ (M1)

$t = 1.18879 \dots$ and $2.23598 \dots$ OR $4.33038 \dots$ and $5.37758 \dots$ OR $7.47197 \dots$ and $8.51917 \dots$ (A1)

Note: Award full marks for $t = \frac{4 π}{3} - 3, \frac{5 π}{3} - 3, (\frac{7 π}{3} - 3, \frac{8 π}{3} - 3 \frac{10 π}{3} - 3, \frac{11 π}{3} - 3)$ .

Award subsequent marks for correct use of these exact values.

$1.18879 \dots < t < 2.23598 \dots$ OR $4.33038 \dots < t < 5.37758 \dots$ OR

$7.47197 \dots < t < 8.51917 \dots$ (A1)

attempts to calculate the total amount of time (M1)

$3 (2.2359 \dots - 1.1887 \dots) (= 3 ((\frac{5 π}{3} - 3) - (\frac{4 π}{3} - 3)))$

$= 3.14 (= π)$ (weeks) A1

[6 marks]

Examiners report

Many students did not change their calculators back to radian mode. This meant they had no chance of correctly answering parts (c) and (d), since even if follow through was given, there were not enough intersections on the graphs.

Most managed part (a) and some attempted to equate the functions in part b) but few recognised that 'rate of growth' was the derivatives of the given functions, and of those who did, most were unable to find them.

Almost all the candidates who did solve part (c) gave the answer $3 \times 1.05 = 3.15$ , when working with more significant figures would have given them 3.14. They lost the last mark.

a.i.

[N/A]

a.ii.

[N/A]

22M.2.SL.TZ1.8c.ii

Question

The function $f$ is defined by $f (x) = \frac{4 x + 1}{x + 4}$ , where $x \in ℝ, x \neq - 4$ .

For the graph of $f$

The graphs of $f$ and $f^{- 1}$ intersect at $x = p$ and $x = q$ , where $p < q$ .

write down the equation of the vertical asymptote.

[1]

a.i.

find the equation of the horizontal asymptote.

[2]

a.ii.

Find $f^{- 1} (x)$ .

[4]

b.i.

Using an algebraic approach, show that the graph of $f^{- 1}$ is obtained by a reflection of the graph of $f$ in the $y$ -axis followed by a reflection in the $x$ -axis.

[4]

b.ii.

Find the value of $p$ and the value of $q$ .

[2]

c.i.

Hence, find the area enclosed by the graph of $f$ and the graph of $f^{- 1}$ .

[3]

c.ii.

Markscheme

$x = - 4$ A1

[1 mark]

a.i.

attempt to substitute into $y = \frac{a}{c}$ OR table with large values of $x$ OR sketch of $f$ showing asymptotic behaviour (M1)

$y = 4$ A1

[2 marks]

a.ii.

$y = \frac{4 x + 1}{x + 4}$

attempt to interchange $x$ and $y$ (seen anywhere) M1

$x y + 4 y = 4 x + 1$ OR $x y + 4 x = 4 y + 1$ (A1)

$x y - 4 x = 1 - 4 y$ OR $x y - 4 y = 1 - 4 x$ (A1)

$f^{- 1} (x) = \frac{1 - 4 x}{x - 4}$ (accept $y = \frac{1 - 4 x}{x - 4}$ ) A1

[4 marks]

b.i.

reflection in $y$ -axis given by $f (- x)$ (M1)

$f (- x) = \frac{- 4 x + 1}{- x + 4}$ (A1)

reflection of their $f (- x)$ in $x$ -axis given by $- f (- x)$ accept "now $- f (x)$ " M1

$(- f (- x) =) - \frac{- 4 x + 1}{- x + 4}$

$= \frac{- 4 x + 1}{x - 4}$ OR $\frac{4 x - 1}{- x + 4}$ A1

$= \frac{1 - 4 x}{x - 4} (= f^{- 1} (x))$ AG

Note: If the candidate attempts to show the result using a particular coordinate on the graph of $f$ rather than a general coordinate on the graph of $f$ , where appropriate, award marks as follows:
M0A0 for eg $(2, 3) \to (- 2, 3)$
M0A0 for $(- 2, 3) \to (- 2, - 3)$

[4 marks]

b.ii.

attempt to solve $f (x) = f^{- 1} (x)$ using graph or algebraically (M1)

$p = - 1$ AND $q = 1$ A1

Note: Award (M1)A0 if only one correct value seen.

[2 marks]

c.i.

attempt to set up an integral to find area between $f$ and $f^{- 1}$ (M1)

$\int_{- 1}^{1} (\frac{4 x + 1}{x + 4} - \frac{1 - 4 x}{x - 4}) d x$ (A1)

$= 0.675231 \dots$

$= 0.675$ A1

[3 marks]

c.ii.

Examiners report

Candidates mostly found the first part of this question accessible, with many knowing how to find the equation of both asymptotes. Common errors included transposing the asymptotes, or finding where an asymptote occurred but not giving it as an equation.

Candidates knew how to start part (b)(i), with most attempting to find the inverse function by firstly interchanging $x$ and $y$ . However, many struggled with the algebra required to change the subject, and were not awarded all the marks. A common error in this part was for candidates to attempt to find an expression for $f' (x)$ , rather than one for $f^{- 1} (x)$ . Few candidates were able to answer part (b)(ii). Many appeared not to know that a reflection in the $y$ -axis is given by $f (- x)$ , or that a reflection in the $x$ -axis is given by $- f (x)$ . Many of those that did, multiplied both the numerator and denominator by $- 1$ when taking the negative of their $f (- x)$ , i.e. $- (\frac{- 4 x + 1}{- x + 4})$ was often simplified as $\frac{4 x - 1}{x - 4}$ . However, the majority of candidates either did not attempt this question part or attempted to describe a graphical approach often involving a specific point, rather than an algebraic approach.

Those that attempted part (c), and had the correct expression for $f^{- 1} (x)$ , were usually able to gain all the marks. However, those that had an incorrect expression, or had found $f' (x)$ , often proceeded to find an area, even when there was not an area enclosed by their two curves.

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

c.i.

[N/A]

c.ii.

EXN.2.SL.TZ0.3a

Question

A particle moves in a straight line such that its velocity, $v {ms}^{- 1}$ , at time $t$ seconds is given by

$v = 4 t^{2} - 6 t + 9 - 2 \sin (4 t), 0 \leq t \leq 1$ .

The particle’s acceleration is zero at $t = T$ .

Find the value of $T$ .

[2]

Let $s_{1}$ be the distance travelled by the particle from $t = 0$ to $t = T$ and let $s_{2}$ be the distance travelled by the particle from $t = T$ to $t = 1$ .

Show that $s_{2} > s_{1}$ .

[3]

Markscheme

attempts either graphical or symbolic means to find the value of $t$ when $\frac{d v}{d t} = 0$ (M1)

$T = 0.465 (s)$ A1

[2 marks]

attempts to find the value of either $s_{1} = \int_{0}^{0.46494 \dots} v d t$ or $s_{2} = \int_{0.46494 \dots}^{1} v d t$ (M1)

$s_{1} = 3.02758 \dots$ and $s_{2} = 3.47892 \dots$ A1A1

Note: Award as above for obtaining, for example, $s_{2} - s_{1} = 0.45133 \dots$ or $\frac{s_{2}}{s_{1}} = 1.14907 \dots$

Note: Award a maximum of M1A1A0FT for use of an incorrect value of $T$ from part (a).

so $s_{2} > s_{1}$ AG

[3 marks]

Examiners report

[N/A]

EXN.2.SL.TZ0.6b

Question

Consider the curves $y = x^{2} \sin x$ and $y = - 1 - \sqrt{1 + 4 {(x + 2)}^{2}}$ for $- π \leq x \leq 0$ .

Find the $x$ -coordinates of the points of intersection of the two curves.

[3]

Find the area, $A$ , of the region enclosed by the two curves.

[4]

Markscheme

attempts to solve $x^{2} \sin x = - 1 - \sqrt{1 + 4 {(x + 2)}^{2}}$ (M1)

$x = - 2.76, - 1.54$ A1A1

Note: Award A1A0 if additional solutions outside the domain are given.

[3 marks]

$A = \int_{- 2.762 \dots}^{- 1.537 \dots} (- 1 - \sqrt{1 + 4 {(x + 2)}^{2}} - x^{2} \sin x) d x$ (or equivalent) (M1)(A1)

Note: Award M1 for attempting to form an integrand involving “top curve” − “bottom curve”.

so $A = 1.47$ A2

[4 marks]

Examiners report

[N/A]

SPM.2.SL.TZ0.6a

Question

The displacement, in centimetres, of a particle from an origin, O, at time t seconds, is given by s(t) = t ² cos t + 2t sin t, 0 ≤ t ≤ 5.

Find the maximum distance of the particle from O.

[3]

Find the acceleration of the particle at the instant it first changes direction.

[4]

Markscheme

use of a graph to find the coordinates of the local minimum (M1)

s = −16.513... (A1)

maximum distance is 16.5 cm (to the left of O) A1

[3 marks]

attempt to find time when particle changes direction eg considering the first maximum on the graph of s or the first t – intercept on the graph of s'. (M1)

t = 1.51986... (A1)

attempt to find the gradient of s' for their value of t, s" (1.51986...) (M1)

=–8.92 (cm/s²) A1

[4 marks]

Examiners report

[N/A]

DP Mathematics: Analysis and Approaches Questionbank

Topic 5 —Calculus

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